#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int maxL = 5;
const int inf = 1e9;
int n, Q;
char s[maxn];
int L = 5;
struct Matrix {
  int x[maxL][maxL];
} I, INF, A, tr[maxn << 2];
Matrix operator*(const Matrix a, const Matrix b) {
  Matrix ret = INF;
  for (int i = 0; i < L; i++)
    for (int j = 0; j < L; j++)
      for (int k = 0; k < L; k++)
        ret.x[i][j] = min(ret.x[i][j], a.x[i][k] + b.x[k][j]);
  return ret;
}
void build(int rt = 1, int l = 1, int r = n) {
  if (l == r) {
    tr[rt] = INF;
    for (int i = 0; i < L; i++) tr[rt].x[i][i] = 0;
    if (s[l] == '2') tr[rt].x[0][0] = 1, tr[rt].x[0][1] = 0;
    if (s[l] == '0') tr[rt].x[1][1] = 1, tr[rt].x[1][2] = 0;
    if (s[l] == '1') tr[rt].x[2][2] = 1, tr[rt].x[2][3] = 0;
    if (s[l] == '7') tr[rt].x[3][3] = 1, tr[rt].x[3][4] = 0;
    if (s[l] == '6') tr[rt].x[3][3] = 1, tr[rt].x[4][4] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  tr[rt] = tr[rt << 1] * tr[rt << 1 | 1];
}
void qry(int nl, int nr, int rt = 1, int l = 1, int r = n) {
  if (nl <= l && nr >= r) {
    A = A * tr[rt];
    return;
  }
  int mid = l + r >> 1;
  if (mid >= nl) qry(nl, nr, rt << 1, l, mid);
  if (mid + 1 <= nr) qry(nl, nr, rt << 1 | 1, mid + 1, r);
  return;
}
int main() {
  for (int i = 0; i < L; i++)
    for (int j = 0; j < L; j++) INF.x[i][j] = inf;
  scanf("%d %d %s", &n, &Q, s + 1);
  build();
  while (Q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    A = INF;
    A.x[0][0] = 0;
    qry(u, v);
    if (A.x[0][4] == inf)
      puts("-1");
    else
      printf("%d\n", A.x[0][4]);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 2e5 + 10;
const int maxL = 5;
const int inf = 1e9;
int n, Q;
char s[maxn];
int L = 5;
struct Matrix {
  int x[maxL][maxL];
} I, INF, A, tr[maxn << 2];
Matrix operator*(const Matrix a, const Matrix b) {
  Matrix ret = INF;
  for (int i = 0; i < L; i++)
    for (int j = 0; j < L; j++)
      for (int k = 0; k < L; k++)
        ret.x[i][j] = min(ret.x[i][j], a.x[i][k] + b.x[k][j]);
  return ret;
}
void build(int rt = 1, int l = 1, int r = n) {
  if (l == r) {
    tr[rt] = INF;
    for (int i = 0; i < L; i++) tr[rt].x[i][i] = 0;
    if (s[l] == '2') tr[rt].x[0][0] = 1, tr[rt].x[0][1] = 0;
    if (s[l] == '0') tr[rt].x[1][1] = 1, tr[rt].x[1][2] = 0;
    if (s[l] == '1') tr[rt].x[2][2] = 1, tr[rt].x[2][3] = 0;
    if (s[l] == '7') tr[rt].x[3][3] = 1, tr[rt].x[3][4] = 0;
    if (s[l] == '6') tr[rt].x[3][3] = 1, tr[rt].x[4][4] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  tr[rt] = tr[rt << 1] * tr[rt << 1 | 1];
}
void qry(int nl, int nr, int rt = 1, int l = 1, int r = n) {
  if (nl <= l && nr >= r) {
    A = A * tr[rt];
    return;
  }
  int mid = l + r >> 1;
  if (mid >= nl) qry(nl, nr, rt << 1, l, mid);
  if (mid + 1 <= nr) qry(nl, nr, rt << 1 | 1, mid + 1, r);
  return;
}
int main() {
  for (int i = 0; i < L; i++)
    for (int j = 0; j < L; j++) INF.x[i][j] = inf;
  scanf("%d %d %s", &n, &Q, s + 1);
  build();
  while (Q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    A = INF;
    A.x[0][0] = 0;
    qry(u, v);
    if (A.x[0][4] == inf)
      puts("-1");
    else
      printf("%d\n", A.x[0][4]);
  }
  return 0;
}

```