#include <bits/stdc++.h>
using namespace std;
int N, M;
char sir[200009];
int ord[] = {2, 0, 1, 7};
struct state {
  int v[5][5];
} aint[800009], ansQ;
state operator+(state a, state b) {
  state c;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      c.v[i][j] = N + 3;
      for (int k = i; k <= j; k++)
        if (a.v[i][k] + b.v[k][j] < c.v[i][j])
          c.v[i][j] = a.v[i][k] + b.v[k][j];
    }
  return c;
}
void Build(int nod, int st, int dr) {
  if (st == dr) {
    int val = sir[st] - '0';
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++) aint[nod].v[i][j] = N + 3;
    for (int i = 0; i < 5; i++) {
      if (i < 4 && val == ord[i]) {
        aint[nod].v[i][i] = 1;
        aint[nod].v[i][i + 1] = 0;
      } else
        aint[nod].v[i][i] = (i >= 3 && val == 6);
    }
    return;
  }
  int mij = (st + dr) >> 1;
  Build(nod << 1, st, mij);
  Build(nod << 1 | 1, mij + 1, dr);
  aint[nod] = aint[nod << 1] + aint[nod << 1 | 1];
}
void Query(int nod, int st, int dr, int x, int y) {
  if (x <= st && dr <= y) {
    if (x == st)
      ansQ = aint[nod];
    else
      ansQ = ansQ + aint[nod];
    return;
  }
  int mij = (st + dr) >> 1;
  if (x <= mij) Query(nod << 1, st, mij, x, y);
  if (mij < y) Query(nod << 1 | 1, mij + 1, dr, x, y);
}
int main() {
  scanf("%d %d\n", &N, &M);
  gets(sir + 1);
  Build(1, 1, N);
  while (M--) {
    int x, y;
    scanf("%d %d", &x, &y), Query(1, 1, N, x, y);
    if (ansQ.v[0][4] > N)
      printf("-1\n");
    else
      printf("%d\n", ansQ.v[0][4]);
  }
  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, M;
char sir[200009];
int ord[] = {2, 0, 1, 7};
struct state {
  int v[5][5];
} aint[800009], ansQ;
state operator+(state a, state b) {
  state c;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      c.v[i][j] = N + 3;
      for (int k = i; k <= j; k++)
        if (a.v[i][k] + b.v[k][j] < c.v[i][j])
          c.v[i][j] = a.v[i][k] + b.v[k][j];
    }
  return c;
}
void Build(int nod, int st, int dr) {
  if (st == dr) {
    int val = sir[st] - '0';
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++) aint[nod].v[i][j] = N + 3;
    for (int i = 0; i < 5; i++) {
      if (i < 4 && val == ord[i]) {
        aint[nod].v[i][i] = 1;
        aint[nod].v[i][i + 1] = 0;
      } else
        aint[nod].v[i][i] = (i >= 3 && val == 6);
    }
    return;
  }
  int mij = (st + dr) >> 1;
  Build(nod << 1, st, mij);
  Build(nod << 1 | 1, mij + 1, dr);
  aint[nod] = aint[nod << 1] + aint[nod << 1 | 1];
}
void Query(int nod, int st, int dr, int x, int y) {
  if (x <= st && dr <= y) {
    if (x == st)
      ansQ = aint[nod];
    else
      ansQ = ansQ + aint[nod];
    return;
  }
  int mij = (st + dr) >> 1;
  if (x <= mij) Query(nod << 1, st, mij, x, y);
  if (mij < y) Query(nod << 1 | 1, mij + 1, dr, x, y);
}
int main() {
  scanf("%d %d\n", &N, &M);
  gets(sir + 1);
  Build(1, 1, N);
  while (M--) {
    int x, y;
    scanf("%d %d", &x, &y), Query(1, 1, N, x, y);
    if (ansQ.v[0][4] > N)
      printf("-1\n");
    else
      printf("%d\n", ansQ.v[0][4]);
  }
  return 0;
}

```