#include <bits/stdc++.h>
using namespace std;
struct node {
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++) p[i][j] = (i == j ? 0 : 1e9);
  }
  int p[5][5];
  node operator*(const node &a) const {
    node res;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        res.p[i][j] = 1e9;
        for (int k = i; k <= j; k++) {
          res.p[i][j] = min(res.p[i][j], p[i][k] + a.p[k][j]);
        }
      }
    }
    return res;
  }
};
node T[1 << 19];
int n, q;
char A[200020];
const int add = 1 << 18;
int last_7[200020];
int get(int x, int y) {
  x += add;
  y += add;
  node resL = node(), resR = node();
  while (x <= y) {
    if (x & 1) resL = resL * T[x];
    if (!(y & 1)) resR = T[y] * resR;
    x = (x + 1) >> 1;
    y = (y - 1) >> 1;
  }
  return (resL * resR).p[0][3];
}
int S[200020];
void solve() {
  scanf("%d%d", &n, &q);
  scanf("%s", A + 1);
  last_7[0] = -1;
  for (int i = 1; i <= n; i++) {
    last_7[i] = last_7[i - 1];
    if (A[i] == '7') last_7[i] = i;
    S[i] = S[i - 1];
    if (A[i] == '6') S[i]++;
  }
  for (int i = 1; i <= n; i++) {
    if (A[i] == '2') {
      T[add + i].p[0][0] = 1;
      T[add + i].p[0][1] = 0;
    } else if (A[i] == '0') {
      T[add + i].p[1][1] = 1;
      T[add + i].p[1][2] = 0;
    } else if (A[i] == '1') {
      T[add + i].p[2][2] = 1;
      T[add + i].p[2][3] = 0;
    } else if (A[i] == '6') {
      T[add + i].p[3][3] = 1;
      T[add + i].p[3][4] = 0;
    }
  }
  for (int i = add - 1; i >= 1; i--) T[i] = T[i + i] * T[i + i + 1];
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (last_7[y] == -1) {
      puts("-1");
      continue;
    }
    int ty = last_7[y];
    int g = get(x, ty);
    if (g == 1e9) {
      puts("-1");
    } else {
      printf("%d\n", g + S[y] - S[ty - 1]);
    }
  }
}
int main() {
  int Tc = 1;
  for (int tc = 1; tc <= Tc; tc++) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
struct node {
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++) p[i][j] = (i == j ? 0 : 1e9);
  }
  int p[5][5];
  node operator*(const node &a) const {
    node res;
    for (int i = 0; i < 5; i++) {
      for (int j = i; j < 5; j++) {
        res.p[i][j] = 1e9;
        for (int k = i; k <= j; k++) {
          res.p[i][j] = min(res.p[i][j], p[i][k] + a.p[k][j]);
        }
      }
    }
    return res;
  }
};
node T[1 << 19];
int n, q;
char A[200020];
const int add = 1 << 18;
int last_7[200020];
int get(int x, int y) {
  x += add;
  y += add;
  node resL = node(), resR = node();
  while (x <= y) {
    if (x & 1) resL = resL * T[x];
    if (!(y & 1)) resR = T[y] * resR;
    x = (x + 1) >> 1;
    y = (y - 1) >> 1;
  }
  return (resL * resR).p[0][3];
}
int S[200020];
void solve() {
  scanf("%d%d", &n, &q);
  scanf("%s", A + 1);
  last_7[0] = -1;
  for (int i = 1; i <= n; i++) {
    last_7[i] = last_7[i - 1];
    if (A[i] == '7') last_7[i] = i;
    S[i] = S[i - 1];
    if (A[i] == '6') S[i]++;
  }
  for (int i = 1; i <= n; i++) {
    if (A[i] == '2') {
      T[add + i].p[0][0] = 1;
      T[add + i].p[0][1] = 0;
    } else if (A[i] == '0') {
      T[add + i].p[1][1] = 1;
      T[add + i].p[1][2] = 0;
    } else if (A[i] == '1') {
      T[add + i].p[2][2] = 1;
      T[add + i].p[2][3] = 0;
    } else if (A[i] == '6') {
      T[add + i].p[3][3] = 1;
      T[add + i].p[3][4] = 0;
    }
  }
  for (int i = add - 1; i >= 1; i--) T[i] = T[i + i] * T[i + i + 1];
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (last_7[y] == -1) {
      puts("-1");
      continue;
    }
    int ty = last_7[y];
    int g = get(x, ty);
    if (g == 1e9) {
      puts("-1");
    } else {
      printf("%d\n", g + S[y] - S[ty - 1]);
    }
  }
}
int main() {
  int Tc = 1;
  for (int tc = 1; tc <= Tc; tc++) {
    solve();
  }
  return 0;
}

```