#include <bits/stdc++.h>
using namespace std;
int tr[524288][5][5], ans[5], tmp[5], n, q;
char buf[200005];
void qu(int at, int lo, int hi, int rlo, int rhi) {
  if (lo == rlo && hi == rhi) {
    for (int i = 0; i < 5; i++) {
      tmp[i] = 1000000;
      for (int j = 0; j < 5; j++) tmp[i] = min(tmp[i], ans[j] + tr[at][j][i]);
    }
    for (int i = 0; i < 5; i++) ans[i] = tmp[i];
  } else {
    int mid = (lo + hi) >> 1;
    if (rhi <= mid)
      qu(at * 2, lo, mid, rlo, rhi);
    else if (rlo >= mid)
      qu(at * 2 + 1, mid, hi, rlo, rhi);
    else {
      qu(at * 2, lo, mid, rlo, mid);
      qu(at * 2 + 1, mid, hi, mid, rhi);
    }
  }
}
int main() {
  scanf("%d%d%s", &n, &q, buf);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++)
      for (int k = 0; k < 5; k++) tr[i + 262144][j][k] = (j != k) * 1000000;
    switch (buf[i]) {
      case '0':
        tr[i + 262144][1][1] = 1;
        tr[i + 262144][1][2] = 0;
        break;
      case '1':
        tr[i + 262144][2][2] = 1;
        tr[i + 262144][2][3] = 0;
        break;
      case '2':
        tr[i + 262144][0][0] = 1;
        tr[i + 262144][0][1] = 0;
        break;
      case '6':
        tr[i + 262144][3][3] = tr[i + 262144][4][4] = 1;
        break;
      case '7':
        tr[i + 262144][3][3] = 1;
        tr[i + 262144][3][4] = 0;
        break;
    }
  }
  for (int i = 262143; i; i--)
    for (int j = 0; j < 5; j++)
      for (int k = 0; k < 5; k++) {
        tr[i][j][k] = 1000000;
        for (int l = 0; l < 5; l++)
          tr[i][j][k] = min(tr[i][j][k], tr[i * 2][j][l] + tr[i * 2 + 1][l][k]);
      }
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    ans[0] = 0;
    ans[1] = ans[2] = ans[3] = ans[4] = 1000000;
    qu(1, 0, 262144, a - 1, b);
    if (ans[4] == 1000000)
      printf("-1\n");
    else
      printf("%d\n", ans[4]);
  }
  return 0;
}

### Prompt

Generate 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 tr[524288][5][5], ans[5], tmp[5], n, q;
char buf[200005];
void qu(int at, int lo, int hi, int rlo, int rhi) {
  if (lo == rlo && hi == rhi) {
    for (int i = 0; i < 5; i++) {
      tmp[i] = 1000000;
      for (int j = 0; j < 5; j++) tmp[i] = min(tmp[i], ans[j] + tr[at][j][i]);
    }
    for (int i = 0; i < 5; i++) ans[i] = tmp[i];
  } else {
    int mid = (lo + hi) >> 1;
    if (rhi <= mid)
      qu(at * 2, lo, mid, rlo, rhi);
    else if (rlo >= mid)
      qu(at * 2 + 1, mid, hi, rlo, rhi);
    else {
      qu(at * 2, lo, mid, rlo, mid);
      qu(at * 2 + 1, mid, hi, mid, rhi);
    }
  }
}
int main() {
  scanf("%d%d%s", &n, &q, buf);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++)
      for (int k = 0; k < 5; k++) tr[i + 262144][j][k] = (j != k) * 1000000;
    switch (buf[i]) {
      case '0':
        tr[i + 262144][1][1] = 1;
        tr[i + 262144][1][2] = 0;
        break;
      case '1':
        tr[i + 262144][2][2] = 1;
        tr[i + 262144][2][3] = 0;
        break;
      case '2':
        tr[i + 262144][0][0] = 1;
        tr[i + 262144][0][1] = 0;
        break;
      case '6':
        tr[i + 262144][3][3] = tr[i + 262144][4][4] = 1;
        break;
      case '7':
        tr[i + 262144][3][3] = 1;
        tr[i + 262144][3][4] = 0;
        break;
    }
  }
  for (int i = 262143; i; i--)
    for (int j = 0; j < 5; j++)
      for (int k = 0; k < 5; k++) {
        tr[i][j][k] = 1000000;
        for (int l = 0; l < 5; l++)
          tr[i][j][k] = min(tr[i][j][k], tr[i * 2][j][l] + tr[i * 2 + 1][l][k]);
      }
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    ans[0] = 0;
    ans[1] = ans[2] = ans[3] = ans[4] = 1000000;
    qu(1, 0, 262144, a - 1, b);
    if (ans[4] == 1000000)
      printf("-1\n");
    else
      printf("%d\n", ans[4]);
  }
  return 0;
}

```