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Petya likes horse racing very much. Horses numbered from l to r take part in the races. Petya wants to evaluate the probability of victory; for some reason, to do that he needs to know the amount of nearly lucky horses' numbers. A nearly luckynumber is an integer number that has at least two lucky digits the distance between which does not exceed k. Petya learned from some of his mates from Lviv that lucky digits are digits 4 and 7. The distance between the digits is the absolute difference between their positions in the number of a horse. For example, if k = 2, then numbers 412395497, 404, 4070400000070004007 are nearly lucky and numbers 4, 4123954997, 4007000040070004007 are not.

Petya prepared t intervals [l_i, r_i] and invented number k, common for all of them. Your task is to find how many nearly happy numbers there are in each of these segments. Since the answers can be quite large, output them modulo 1000000007 (10⁹ + 7).

Input

The first line contains two integers t and k (1 ≤ t, k ≤ 1000) — the number of segments and the distance between the numbers correspondingly. Next t lines contain pairs of integers l_i and r_i (1 ≤ l ≤ r ≤ 10¹⁰⁰⁰). All numbers are given without the leading zeroes. Numbers in each line are separated by exactly one space character.

Output

Output t lines. In each line print one integer — the answer for the corresponding segment modulo 1000000007 (10⁹ + 7).

Example

Input

1 21 100

Output

4

Input

1 270 77

Output

2

Input

2 11 2080 100

Output

00

Note

In the first sample, the four nearly lucky numbers are 44, 47, 74, 77.

In the second sample, only 74 and 77 are in the given segment.

给定k，求[l,r]内lucky number个数。r的范围到1e1000。1.lucky number是至少存在2个lucky digit的距离小于等于k，并不要求所有都小于等于k。lucky digit是4和7.

思路：

数位dp，两个状态，一是为1代表含有lucky number数，一个是不含，其余难点在那个按位或的上面，没有完全理解

代码

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int  mod=1e9+7;ll t,k;ll bit[1020];ll dp[1020][1020][2];ll dfs(int pos,int pre,int sta,bool limit){if (pos==0){if(sta) return 1;return 0;}if (!limit && dp[pos][pre][sta] != -1) return dp[pos][pre][sta];int up=limit? bit[pos]:9;ll tmp = 0;for (int i = 0; i<= up; i++){if (i ==4||i == 7)tmp+=dfs(pos-1,pos, (pre && (pre - pos) <= k) | sta, limit && i == up);  //符合的情况elsetmp += dfs(pos-1, pre, sta, limit && i == up);            tmp%=mod;}if (!limit)dp[pos][pre][sta] = tmp;return tmp;}ll cal(string s){int pos = 0;for (int i = s.length()-1; i >= 0; i--){bit[++pos]=s[i] - '0';}return dfs(pos,0, 0, 1);}bool check(string s) //用来判断l的，注意是不包含边界，所以单独判断l{int pre = 0;for (int i=1;i<=s.length();i++){if (s[i-1]== '4' ||s[i - 1] == '7'){if (!pre || i - pre > k)pre = i;else if (i-pre <= k)return true;}}return false;}int main(){ios::sync_with_stdio(0);string l, r;cin >> t >> k;memset(dp, -1, sizeof dp);while (t--){cin >> l;cin >> r;ll ans = cal(r)-cal(l)+(check(l) ? 1 : 0);ans=(ans%mod+mod)%mod;cout << ans << endl;}return 0;}