CodeForces 1277 A Happy Birthday, Polycarp!

Description：

Hooray! Polycarp turned n years old! The Technocup Team sincerely congratulates Polycarp!

Polycarp celebrated all of his n birthdays: from the $1-th$ to the $n-th$. At the moment, he is wondering: how many times he turned beautiful number of years?

According to Polycarp, a positive integer is beautiful if it consists of only one digit repeated one or more times. For example, the following numbers are beautiful: $1, 77, 777, 44$ and $999999$. The following numbers are not beautiful: $12, 11110, 6969$and $987654321$

Of course, Polycarpus uses the decimal numeral system ($i.e$. radix is $10$).

Help Polycarpus to find the number of numbers from $1$ to $n$

Input

The first line contains an integer $t (1≤t≤10^4)$— the number of test cases in the input. Then $t$

Each test case consists of one line, which contains a positive integer $n$$(1≤n≤10^9)$— how many years Polycarp has turned.

Output

Print t integers — the answers to the given test cases in the order they are written in the test. Each answer is an integer: the number of beautiful years between $1$ and $n$, inclusive.

Example

input

6
18
1
9
100500
33
1000000000

output

10
1
9
45
12
81

Note

In the first test case of the example beautiful years are $1, 2, 3, 4, 5, 6, 7, 8, 9$and $11$.

题意： 给出$n$让你求前$n$个数中有多少个漂亮数字，就是$111,2222,3333$这样的数字，只有一个数字，前$10$个数里有9个，往后每扩大$10$倍就会多$9$个。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a)
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

int n, t;
int ans, res, cnt, temp;

int main()
{
    sd(t);
    while (t--)
    {
        ans = 0;
        sd(n);
        if (n <= 9)
        {
            cout << n << endl;
            continue;
        }
        int len = log10(n) + 1; //长度
        //cout << len << endl;
        ans = (len - 1) * 9;
        res = n / pow(10, len - 1); //取出最高位数字
        //cout<<res<<endl;
        int sum = 0;
        rep(i, 1, len)
        {
            sum = sum * 10 + res;
        }
        if (n >= sum)
            ans += res;
        else
            ans += res - 1;
        pd(ans);
    }
    return 0;
}