The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <stack>

#include <vector>

#include <queue>

#include <cstring>

#include <string>

#include <map>

#include <set>

using namespace std;

#define LL long long

LL read() {

    LL x = 0, f = 1; char c = getchar();

    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

    return x * f;

}

#define maxn 21

LL f[maxn][maxn], ten[maxn];

int num[maxn], cnt;

LL sum(LL x) {

	cnt = 0; LL orx = x;

	while(x) num[++cnt] = x % 10, x /= 10;

	LL sum = 0;

	for(int i = cnt; i; i--) {

		for(int j = 0; j < num[i]; j++) sum += f[i][j];

		if(i < cnt && num[i+1] == 4 && num[i] == 9) {

			sum += orx % ten[i-1] + 1; break;

		}

	}

	return sum;

}

int main() {

	ten[0] = 1;

	for(int i = 1; i < maxn; i++) ten[i] = ten[i-1] * 10;

	for(int i = 1; i < maxn - 1; i++)

		for(int j = 0; j <= 9; j++)

			for(int k = 0; k <= 9; k++) {

				if(k == 4 && j == 9) f[i+1][k] += ten[i-1];

				else f[i+1][k] += f[i][j];

			}

	int T = read();

	while(T--) {

		LL n = read();

		printf("%lld\n", sum(n));

	}

	return 0;

}