#include <bits/stdc++.h>
using namespace std;
int v[800005][6][6];
int n, q;
char s[200005];
void init(int x, int b, int e) {
  if (b == e) {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[x][i][j] = 1000000000;
    for (int i = 0; i < 6; i++) v[x][i][i] = 0;
    if (s[b] == '2')
      v[x][0][1] = 0, v[x][0][0] = 1;
    else if (s[b] == '0')
      v[x][1][2] = 0, v[x][1][1] = 1;
    else if (s[b] == '1')
      v[x][2][3] = 0, v[x][2][2] = 1;
    else if (s[b] == '7')
      v[x][3][4] = 0, v[x][3][3] = 1;
    else if (s[b] == '6')
      v[x][3][5] = v[x][4][5] = 0, v[x][3][3] = v[x][4][4] = 1;
  } else {
    int l = x * 2, r = x * 2 + 1;
    init(l, b, (((b) + (e)) >> 1));
    init(r, (((b) + (e)) >> 1) + 1, e);
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[x][i][j] = 1000000000;
    for (int k = 0; k < 6; k++)
      for (int i = 0; i < 6; i++)
        for (int j = 0; j < 6; j++)
          v[x][i][j] = min(v[x][i][j], v[l][i][k] + v[r][k][j]);
  }
}
struct mat {
  int v[6][6];
  mat() {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[i][j] = 1000000000;
  }
};
mat merge(mat a, mat b) {
  mat c;
  for (int k = 0; k < 6; k++)
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++)
        c.v[i][j] = min(c.v[i][j], a.v[i][k] + b.v[k][j]);
  return c;
}
mat query(int x, int b, int e, int qb, int qe) {
  if (b > qe || e < qb) return mat();
  if (b >= qb && e <= qe) {
    mat c;
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) c.v[i][j] = v[x][i][j];
    return c;
  } else {
    int m = (((b) + (e)) >> 1);
    if (qe <= m)
      return query(x * 2, b, (((b) + (e)) >> 1), qb, qe);
    else if (qb > m)
      return query(x * 2 + 1, (((b) + (e)) >> 1) + 1, e, qb, qe);
    else
      return merge(query(x * 2, b, (((b) + (e)) >> 1), qb, qe),
                   query(x * 2 + 1, (((b) + (e)) >> 1) + 1, e, qb, qe));
  }
}
int main() {
  scanf("%d%d%s", &n, &q, s);
  int x, y;
  init(1, 0, n - 1);
  while (q--) {
    scanf("%d%d", &x, &y);
    --x, --y;
    mat c = query(1, 0, n - 1, x, y);
    printf("%d\n", (c.v[0][4] >= 1000000000) ? -1 : c.v[0][4]);
  }
}

### 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;
int v[800005][6][6];
int n, q;
char s[200005];
void init(int x, int b, int e) {
  if (b == e) {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[x][i][j] = 1000000000;
    for (int i = 0; i < 6; i++) v[x][i][i] = 0;
    if (s[b] == '2')
      v[x][0][1] = 0, v[x][0][0] = 1;
    else if (s[b] == '0')
      v[x][1][2] = 0, v[x][1][1] = 1;
    else if (s[b] == '1')
      v[x][2][3] = 0, v[x][2][2] = 1;
    else if (s[b] == '7')
      v[x][3][4] = 0, v[x][3][3] = 1;
    else if (s[b] == '6')
      v[x][3][5] = v[x][4][5] = 0, v[x][3][3] = v[x][4][4] = 1;
  } else {
    int l = x * 2, r = x * 2 + 1;
    init(l, b, (((b) + (e)) >> 1));
    init(r, (((b) + (e)) >> 1) + 1, e);
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[x][i][j] = 1000000000;
    for (int k = 0; k < 6; k++)
      for (int i = 0; i < 6; i++)
        for (int j = 0; j < 6; j++)
          v[x][i][j] = min(v[x][i][j], v[l][i][k] + v[r][k][j]);
  }
}
struct mat {
  int v[6][6];
  mat() {
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) v[i][j] = 1000000000;
  }
};
mat merge(mat a, mat b) {
  mat c;
  for (int k = 0; k < 6; k++)
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++)
        c.v[i][j] = min(c.v[i][j], a.v[i][k] + b.v[k][j]);
  return c;
}
mat query(int x, int b, int e, int qb, int qe) {
  if (b > qe || e < qb) return mat();
  if (b >= qb && e <= qe) {
    mat c;
    for (int i = 0; i < 6; i++)
      for (int j = 0; j < 6; j++) c.v[i][j] = v[x][i][j];
    return c;
  } else {
    int m = (((b) + (e)) >> 1);
    if (qe <= m)
      return query(x * 2, b, (((b) + (e)) >> 1), qb, qe);
    else if (qb > m)
      return query(x * 2 + 1, (((b) + (e)) >> 1) + 1, e, qb, qe);
    else
      return merge(query(x * 2, b, (((b) + (e)) >> 1), qb, qe),
                   query(x * 2 + 1, (((b) + (e)) >> 1) + 1, e, qb, qe));
  }
}
int main() {
  scanf("%d%d%s", &n, &q, s);
  int x, y;
  init(1, 0, n - 1);
  while (q--) {
    scanf("%d%d", &x, &y);
    --x, --y;
    mat c = query(1, 0, n - 1, x, y);
    printf("%d\n", (c.v[0][4] >= 1000000000) ? -1 : c.v[0][4]);
  }
}

```