#include <bits/stdc++.h>
using namespace std;
int n, m;
char str[200005];
struct Mat {
  int dp[5][5];
  Mat() { memset(dp, 0x3f3f3f3f, sizeof(dp)); }
} A[200005 << 2];
Mat Merge(const Mat &l, const Mat &r) {
  Mat tmp;
  for (int i = (0), i_end_ = (4); i <= i_end_; i++) {
    for (int j = (i), j_end_ = (4); j <= j_end_; j++) {
      for (int k = (j), k_end_ = (4); k <= k_end_; k++) {
        tmp.dp[i][k] = min(tmp.dp[i][k], l.dp[i][j] + r.dp[j][k]);
      }
    }
  }
  return tmp;
}
void build(int L, int R, int p) {
  if (L == R) {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++) A[p].dp[i][i] = 0;
    if (str[L] == '2') {
      A[p].dp[0][1] = 0;
      A[p].dp[0][0] = 1;
    }
    if (str[L] == '0') {
      A[p].dp[1][2] = 0;
      A[p].dp[1][1] = 1;
    }
    if (str[L] == '1') {
      A[p].dp[2][3] = 0;
      A[p].dp[2][2] = 1;
    }
    if (str[L] == '7') {
      A[p].dp[3][4] = 0;
      A[p].dp[3][3] = 1;
    }
    if (str[L] == '6') {
      A[p].dp[3][3] = 1;
      A[p].dp[4][4] = 1;
    }
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, p << 1);
  build(mid + 1, R, p << 1 | 1);
  A[p] = Merge(A[p << 1], A[p << 1 | 1]);
}
Mat query(int L, int R, int p, int l, int r) {
  if (L == l && R == r) return A[p];
  int mid = (L + R) >> 1;
  if (r <= mid)
    return query(L, mid, p << 1, l, r);
  else if (l > mid)
    return query(mid + 1, R, p << 1 | 1, l, r);
  else
    return Merge(query(L, mid, p << 1, l, mid),
                 query(mid + 1, R, p << 1 | 1, mid + 1, r));
}
int main() {
  scanf("%d%d%s", &n, &m, str + 1);
  build(1, n, 1);
  for (int i = (1), i_end_ = (m); i <= i_end_; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    Mat res = query(1, n, 1, L, R);
    printf("%d\n", res.dp[0][4] == 0x3f3f3f3f ? -1 : res.dp[0][4]);
  }
  return 0;
}

### 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 n, m;
char str[200005];
struct Mat {
  int dp[5][5];
  Mat() { memset(dp, 0x3f3f3f3f, sizeof(dp)); }
} A[200005 << 2];
Mat Merge(const Mat &l, const Mat &r) {
  Mat tmp;
  for (int i = (0), i_end_ = (4); i <= i_end_; i++) {
    for (int j = (i), j_end_ = (4); j <= j_end_; j++) {
      for (int k = (j), k_end_ = (4); k <= k_end_; k++) {
        tmp.dp[i][k] = min(tmp.dp[i][k], l.dp[i][j] + r.dp[j][k]);
      }
    }
  }
  return tmp;
}
void build(int L, int R, int p) {
  if (L == R) {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++) A[p].dp[i][i] = 0;
    if (str[L] == '2') {
      A[p].dp[0][1] = 0;
      A[p].dp[0][0] = 1;
    }
    if (str[L] == '0') {
      A[p].dp[1][2] = 0;
      A[p].dp[1][1] = 1;
    }
    if (str[L] == '1') {
      A[p].dp[2][3] = 0;
      A[p].dp[2][2] = 1;
    }
    if (str[L] == '7') {
      A[p].dp[3][4] = 0;
      A[p].dp[3][3] = 1;
    }
    if (str[L] == '6') {
      A[p].dp[3][3] = 1;
      A[p].dp[4][4] = 1;
    }
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, p << 1);
  build(mid + 1, R, p << 1 | 1);
  A[p] = Merge(A[p << 1], A[p << 1 | 1]);
}
Mat query(int L, int R, int p, int l, int r) {
  if (L == l && R == r) return A[p];
  int mid = (L + R) >> 1;
  if (r <= mid)
    return query(L, mid, p << 1, l, r);
  else if (l > mid)
    return query(mid + 1, R, p << 1 | 1, l, r);
  else
    return Merge(query(L, mid, p << 1, l, mid),
                 query(mid + 1, R, p << 1 | 1, mid + 1, r));
}
int main() {
  scanf("%d%d%s", &n, &m, str + 1);
  build(1, n, 1);
  for (int i = (1), i_end_ = (m); i <= i_end_; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    Mat res = query(1, n, 1, L, R);
    printf("%d\n", res.dp[0][4] == 0x3f3f3f3f ? -1 : res.dp[0][4]);
  }
  return 0;
}

```