#include <bits/stdc++.h>
using namespace std;
int n, q;
char wcz[1000007];
int tab[1000007];
int n1;
int inf = 1000000;
int drzewo[530000][5][5];
int sta[5];
int now[5];
int potenga(int v) {
  for (int i = 1; 1; i <<= 1)
    if (i >= v) return i;
}
void start(int v, int a, int b) {
  for (int i = 0; i <= 4; i++)
    for (int j = 0; j <= 4; j++) drzewo[v][i][j] = inf;
  if (a == b) {
    if (!tab[a]) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
    }
    if (tab[a] == 1) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][0][0] = 1;
      drzewo[v][0][1] = 0;
    }
    if (tab[a] == 2) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][1][1] = 1;
      drzewo[v][1][2] = 0;
    }
    if (tab[a] == 3) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][2][2] = 1;
      drzewo[v][2][3] = 0;
    }
    if (tab[a] == 4) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][3][3] = 1;
      drzewo[v][3][4] = 0;
    }
    if (tab[a] == -1) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][3][3] = 1;
      drzewo[v][4][4] = 1;
    }
    return;
  }
  start((v << 1), a, (a + b) >> 1);
  start((v << 1) ^ 1, (a + b + 2) >> 1, b);
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++)
      for (int l = j; l <= 4; l++)
        drzewo[v][i][l] = min(drzewo[v][i][l], drzewo[(v << 1)][i][j] +
                                                   drzewo[(v << 1) ^ 1][j][l]);
}
void czyt(int v, int a, int b, int graa, int grab) {
  if (a >= graa && b <= grab) {
    for (int i = 0; i <= 4; i++) {
      now[i] = sta[i];
      sta[i] = inf;
    }
    for (int i = 0; i <= 4; i++)
      for (int j = i; j <= 4; j++)
        sta[j] = min(sta[j], now[i] + drzewo[v][i][j]);
    return;
  }
  if (a > grab || b < graa) {
    return;
  }
  czyt((v << 1), a, (a + b) >> 1, graa, grab);
  czyt((v << 1) ^ 1, (a + b + 2) >> 1, b, graa, grab);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", wcz + 1);
  n1 = potenga(n);
  for (int i = 1; i <= n; i++) {
    if (wcz[i] == '2') tab[i] = 1;
    if (wcz[i] == '0') tab[i] = 2;
    if (wcz[i] == '1') tab[i] = 3;
    if (wcz[i] == '6') tab[i] = -1;
    if (wcz[i] == '7') tab[i] = 4;
  }
  start(1, 1, n1);
  while (q--) {
    int p1, p2;
    scanf("%d%d", &p1, &p2);
    sta[1] = sta[2] = sta[3] = sta[4] = inf;
    sta[0] = 0;
    czyt(1, 1, n1, p1, p2);
    if (sta[4] < inf)
      printf("%d\n", sta[4]);
    else
      printf("-1\n");
  }
  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, q;
char wcz[1000007];
int tab[1000007];
int n1;
int inf = 1000000;
int drzewo[530000][5][5];
int sta[5];
int now[5];
int potenga(int v) {
  for (int i = 1; 1; i <<= 1)
    if (i >= v) return i;
}
void start(int v, int a, int b) {
  for (int i = 0; i <= 4; i++)
    for (int j = 0; j <= 4; j++) drzewo[v][i][j] = inf;
  if (a == b) {
    if (!tab[a]) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
    }
    if (tab[a] == 1) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][0][0] = 1;
      drzewo[v][0][1] = 0;
    }
    if (tab[a] == 2) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][1][1] = 1;
      drzewo[v][1][2] = 0;
    }
    if (tab[a] == 3) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][2][2] = 1;
      drzewo[v][2][3] = 0;
    }
    if (tab[a] == 4) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][3][3] = 1;
      drzewo[v][3][4] = 0;
    }
    if (tab[a] == -1) {
      for (int i = 0; i <= 4; i++) drzewo[v][i][i] = 0;
      drzewo[v][3][3] = 1;
      drzewo[v][4][4] = 1;
    }
    return;
  }
  start((v << 1), a, (a + b) >> 1);
  start((v << 1) ^ 1, (a + b + 2) >> 1, b);
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++)
      for (int l = j; l <= 4; l++)
        drzewo[v][i][l] = min(drzewo[v][i][l], drzewo[(v << 1)][i][j] +
                                                   drzewo[(v << 1) ^ 1][j][l]);
}
void czyt(int v, int a, int b, int graa, int grab) {
  if (a >= graa && b <= grab) {
    for (int i = 0; i <= 4; i++) {
      now[i] = sta[i];
      sta[i] = inf;
    }
    for (int i = 0; i <= 4; i++)
      for (int j = i; j <= 4; j++)
        sta[j] = min(sta[j], now[i] + drzewo[v][i][j]);
    return;
  }
  if (a > grab || b < graa) {
    return;
  }
  czyt((v << 1), a, (a + b) >> 1, graa, grab);
  czyt((v << 1) ^ 1, (a + b + 2) >> 1, b, graa, grab);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", wcz + 1);
  n1 = potenga(n);
  for (int i = 1; i <= n; i++) {
    if (wcz[i] == '2') tab[i] = 1;
    if (wcz[i] == '0') tab[i] = 2;
    if (wcz[i] == '1') tab[i] = 3;
    if (wcz[i] == '6') tab[i] = -1;
    if (wcz[i] == '7') tab[i] = 4;
  }
  start(1, 1, n1);
  while (q--) {
    int p1, p2;
    scanf("%d%d", &p1, &p2);
    sta[1] = sta[2] = sta[3] = sta[4] = inf;
    sta[0] = 0;
    czyt(1, 1, n1, p1, p2);
    if (sta[4] < inf)
      printf("%d\n", sta[4]);
    else
      printf("-1\n");
  }
  return 0;
}

```