#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
struct query {
  int s, e, idx;
};
int n, q, ret[MAXN];
char str[MAXN];
int dl[MAXN][6], dr[MAXN][6];
int code(char c) {
  if (c == '2') return 0;
  if (c == '0') return 1;
  if (c == '1') return 2;
  if (c == '6') return 3;
  if (c == '7') return 4;
  return -1;
}
void solve(int s, int e, vector<query> v) {
  if (s == e) {
    return;
  }
  int m = (s + e) / 2;
  vector<query> l, r, x;
  for (auto &i : v) {
    if (i.e <= m)
      l.push_back(i);
    else if (i.s > m)
      r.push_back(i);
    else
      x.push_back(i);
  }
  solve(s, m, l);
  solve(m + 1, e, r);
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 6; j++) {
      dl[m + 1][j] = dr[m][j] = 1e9;
    }
    dl[m + 1][i] = dr[m][i] = 0;
    for (int j = m; j >= s; j--) {
      int val = code(str[j]);
      for (int k = 0; k < 6; k++) {
        if (val == -1) {
          dl[j][k] = dl[j + 1][k];
        } else if (val <= 2) {
          if (k == val)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][k + 1]);
          else
            dl[j][k] = dl[j + 1][k];
        } else if (val == 3) {
          if (k == 3 || k == 4)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][5]);
          else
            dl[j][k] = dl[j + 1][k];
        } else {
          if (k == 3)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][k + 1]);
          else
            dl[j][k] = dl[j + 1][k];
        }
      }
    }
    for (int j = m + 1; j <= e; j++) {
      int val = code(str[j]);
      for (int k = 0; k < 6; k++) {
        if (val == -1) {
          dr[j][k] = dr[j - 1][k];
        } else if (val <= 2) {
          if (k == val + 1)
            dr[j][k] = min(dr[j - 1][k], dr[j - 1][k - 1]);
          else
            dr[j][k] = dr[j - 1][k] + (k == val);
        } else if (val == 3) {
          if (k == 3 || k == 4)
            dr[j][k] = dr[j - 1][k] + 1;
          else if (k == 5)
            dr[j][k] = min({dr[j - 1][3], dr[j - 1][4], dr[j - 1][5]});
          else
            dr[j][k] = dr[j - 1][k];
        } else {
          if (k == 4)
            dr[j][k] = min(dr[j - 1][k], dr[j - 1][k - 1]);
          else
            dr[j][k] = dr[j - 1][k] + (k == 3);
        }
      }
    }
    for (auto &i : x) {
      ret[i.idx] = min(dl[i.s][0] + dr[i.e][4], ret[i.idx]);
    }
  }
}
int main() {
  scanf("%d %d %s", &n, &q, str + 1);
  vector<query> v;
  for (int i = 0; i < q; i++) {
    query qr;
    scanf("%d %d", &qr.s, &qr.e);
    qr.idx = i;
    v.push_back(qr);
  }
  memset(ret, 0x3f, sizeof(ret));
  solve(1, n, v);
  for (int i = 0; i < q; i++) {
    if (ret[i] > 1e8) ret[i] = -1;
    printf("%d\n", ret[i]);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int MAXN = 200005;
struct query {
  int s, e, idx;
};
int n, q, ret[MAXN];
char str[MAXN];
int dl[MAXN][6], dr[MAXN][6];
int code(char c) {
  if (c == '2') return 0;
  if (c == '0') return 1;
  if (c == '1') return 2;
  if (c == '6') return 3;
  if (c == '7') return 4;
  return -1;
}
void solve(int s, int e, vector<query> v) {
  if (s == e) {
    return;
  }
  int m = (s + e) / 2;
  vector<query> l, r, x;
  for (auto &i : v) {
    if (i.e <= m)
      l.push_back(i);
    else if (i.s > m)
      r.push_back(i);
    else
      x.push_back(i);
  }
  solve(s, m, l);
  solve(m + 1, e, r);
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 6; j++) {
      dl[m + 1][j] = dr[m][j] = 1e9;
    }
    dl[m + 1][i] = dr[m][i] = 0;
    for (int j = m; j >= s; j--) {
      int val = code(str[j]);
      for (int k = 0; k < 6; k++) {
        if (val == -1) {
          dl[j][k] = dl[j + 1][k];
        } else if (val <= 2) {
          if (k == val)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][k + 1]);
          else
            dl[j][k] = dl[j + 1][k];
        } else if (val == 3) {
          if (k == 3 || k == 4)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][5]);
          else
            dl[j][k] = dl[j + 1][k];
        } else {
          if (k == 3)
            dl[j][k] = min(dl[j + 1][k] + 1, dl[j + 1][k + 1]);
          else
            dl[j][k] = dl[j + 1][k];
        }
      }
    }
    for (int j = m + 1; j <= e; j++) {
      int val = code(str[j]);
      for (int k = 0; k < 6; k++) {
        if (val == -1) {
          dr[j][k] = dr[j - 1][k];
        } else if (val <= 2) {
          if (k == val + 1)
            dr[j][k] = min(dr[j - 1][k], dr[j - 1][k - 1]);
          else
            dr[j][k] = dr[j - 1][k] + (k == val);
        } else if (val == 3) {
          if (k == 3 || k == 4)
            dr[j][k] = dr[j - 1][k] + 1;
          else if (k == 5)
            dr[j][k] = min({dr[j - 1][3], dr[j - 1][4], dr[j - 1][5]});
          else
            dr[j][k] = dr[j - 1][k];
        } else {
          if (k == 4)
            dr[j][k] = min(dr[j - 1][k], dr[j - 1][k - 1]);
          else
            dr[j][k] = dr[j - 1][k] + (k == 3);
        }
      }
    }
    for (auto &i : x) {
      ret[i.idx] = min(dl[i.s][0] + dr[i.e][4], ret[i.idx]);
    }
  }
}
int main() {
  scanf("%d %d %s", &n, &q, str + 1);
  vector<query> v;
  for (int i = 0; i < q; i++) {
    query qr;
    scanf("%d %d", &qr.s, &qr.e);
    qr.idx = i;
    v.push_back(qr);
  }
  memset(ret, 0x3f, sizeof(ret));
  solve(1, n, v);
  for (int i = 0; i < q; i++) {
    if (ret[i] > 1e8) ret[i] = -1;
    printf("%d\n", ret[i]);
  }
}

```