#include <bits/stdc++.h>
using namespace std;
int n, q;
const int N = 2e5 + 10;
const int inf = 1e9;
char s[N];
struct Node {
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = (i == j) ? 0 : inf;
      }
    }
  }
  int dp[5][5];
};
Node operator*(const Node& u, const Node& v) {
  Node ans;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      ans.dp[i][j] = inf;
      for (int k = 0; k < 5; k++) {
        ans.dp[i][j] = min(u.dp[i][k] + v.dp[k][j], ans.dp[i][j]);
      }
    }
  }
  return ans;
}
Node nodes[N * 4];
void build_tree(int l, int r, int rt) {
  Node& v = nodes[rt];
  if (l == r) {
    char c = s[l - 1];
    if (c == '2') {
      v.dp[0][0] = 1;
      v.dp[0][1] = 0;
    } else if (c == '0') {
      v.dp[1][1] = 1;
      v.dp[1][2] = 0;
    } else if (c == '1') {
      v.dp[2][2] = 1;
      v.dp[2][3] = 0;
    } else if (c == '7') {
      v.dp[3][3] = 1;
      v.dp[3][4] = 0;
    } else if (c == '6') {
      v.dp[3][3] = 1;
      v.dp[4][4] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build_tree(l, m, 2 * rt);
  build_tree(m + 1, r, 2 * rt + 1);
  v = nodes[2 * rt] * nodes[2 * rt + 1];
}
Node query(int L, int R, int l, int r, int rt) {
  Node& v = nodes[rt];
  if (L <= l && R >= r) return v;
  int m = (l + r) >> 1;
  bool left = false, right = false;
  Node ltemp, rtemp;
  if (L <= m) {
    ltemp = query(L, R, l, m, rt * 2);
    left = true;
  }
  if (R > m) {
    rtemp = query(L, R, m + 1, r, rt * 2 + 1);
    right = true;
  }
  if (left && right) return ltemp * rtemp;
  if (left) return ltemp;
  if (right) return rtemp;
}
int main() {
  scanf("%d %d", &n, &q);
  scanf("%s", s);
  build_tree(1, n, 1);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    Node v = query(l, r, 1, n, 1);
    if (v.dp[0][4] == inf)
      cout << -1 << endl;
    else
      cout << v.dp[0][4] << endl;
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, q;
const int N = 2e5 + 10;
const int inf = 1e9;
char s[N];
struct Node {
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j] = (i == j) ? 0 : inf;
      }
    }
  }
  int dp[5][5];
};
Node operator*(const Node& u, const Node& v) {
  Node ans;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      ans.dp[i][j] = inf;
      for (int k = 0; k < 5; k++) {
        ans.dp[i][j] = min(u.dp[i][k] + v.dp[k][j], ans.dp[i][j]);
      }
    }
  }
  return ans;
}
Node nodes[N * 4];
void build_tree(int l, int r, int rt) {
  Node& v = nodes[rt];
  if (l == r) {
    char c = s[l - 1];
    if (c == '2') {
      v.dp[0][0] = 1;
      v.dp[0][1] = 0;
    } else if (c == '0') {
      v.dp[1][1] = 1;
      v.dp[1][2] = 0;
    } else if (c == '1') {
      v.dp[2][2] = 1;
      v.dp[2][3] = 0;
    } else if (c == '7') {
      v.dp[3][3] = 1;
      v.dp[3][4] = 0;
    } else if (c == '6') {
      v.dp[3][3] = 1;
      v.dp[4][4] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build_tree(l, m, 2 * rt);
  build_tree(m + 1, r, 2 * rt + 1);
  v = nodes[2 * rt] * nodes[2 * rt + 1];
}
Node query(int L, int R, int l, int r, int rt) {
  Node& v = nodes[rt];
  if (L <= l && R >= r) return v;
  int m = (l + r) >> 1;
  bool left = false, right = false;
  Node ltemp, rtemp;
  if (L <= m) {
    ltemp = query(L, R, l, m, rt * 2);
    left = true;
  }
  if (R > m) {
    rtemp = query(L, R, m + 1, r, rt * 2 + 1);
    right = true;
  }
  if (left && right) return ltemp * rtemp;
  if (left) return ltemp;
  if (right) return rtemp;
}
int main() {
  scanf("%d %d", &n, &q);
  scanf("%s", s);
  build_tree(1, n, 1);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    Node v = query(l, r, 1, n, 1);
    if (v.dp[0][4] == inf)
      cout << -1 << endl;
    else
      cout << v.dp[0][4] << endl;
  }
  return 0;
}

```