#include <bits/stdc++.h>
int rd() {
  int k = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') k = k * 10 + c - '0', c = getchar();
  return k;
}
const int N = 200001;
int Min(int x, int y) { return x < y ? x : y; }
struct Mat {
  int a[5][5];
  Mat(int s = 0) {
    memset(a, 0x3f, sizeof(a));
    if (s) a[0][0] = a[1][1] = a[2][2] = a[3][3] = a[4][4] = 0;
  }
};
Mat operator*(Mat x, Mat y) {
  Mat z;
  for (int i = 0; i < 5; ++i)
    for (int j = 0; j < 5; ++j)
      for (int k = 0; k < 5; ++k)
        z.a[i][j] = Min(z.a[i][j], x.a[i][k] + y.a[k][j]);
  return z;
}
Mat Get(int x) {
  Mat g(1);
  if (x == 2) g.a[0][0] = 1, g.a[1][0] = 0;
  if (x == 0) g.a[1][1] = 1, g.a[2][1] = 0;
  if (x == 1) g.a[2][2] = 1, g.a[3][2] = 0;
  if (x == 6) g.a[3][3] = g.a[4][4] = 1;
  if (x == 7) g.a[3][3] = 1, g.a[4][3] = 0;
  return g;
}
int n, q, l, r;
Mat t[N << 2], s;
char c[N];
void Make(int x, int l, int r) {
  if (l == r) {
    t[x] = Get(c[l] - '0');
    return;
  }
  Make((x << 1), l, ((l + r) >> 1)), Make((x << 1 | 1), ((l + r) >> 1) + 1, r),
      t[x] = t[(x << 1 | 1)] * t[(x << 1)];
}
Mat Query(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[x];
  Mat a(1), b(1);
  if (L <= ((l + r) >> 1)) a = Query((x << 1), l, ((l + r) >> 1), L, R);
  if (((l + r) >> 1) < R) b = Query((x << 1 | 1), ((l + r) >> 1) + 1, r, L, R);
  return b * a;
}
int main() {
  n = rd(), q = rd(), scanf("%s", c + 1);
  Make(1, 1, n);
  while (q--) {
    l = rd(), r = rd(), s = Mat(1);
    s = Query(1, 1, n, l, r),
    printf("%d\n", s.a[4][0] > (r - l + 1) ? -1 : s.a[4][0]);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
int rd() {
  int k = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') k = k * 10 + c - '0', c = getchar();
  return k;
}
const int N = 200001;
int Min(int x, int y) { return x < y ? x : y; }
struct Mat {
  int a[5][5];
  Mat(int s = 0) {
    memset(a, 0x3f, sizeof(a));
    if (s) a[0][0] = a[1][1] = a[2][2] = a[3][3] = a[4][4] = 0;
  }
};
Mat operator*(Mat x, Mat y) {
  Mat z;
  for (int i = 0; i < 5; ++i)
    for (int j = 0; j < 5; ++j)
      for (int k = 0; k < 5; ++k)
        z.a[i][j] = Min(z.a[i][j], x.a[i][k] + y.a[k][j]);
  return z;
}
Mat Get(int x) {
  Mat g(1);
  if (x == 2) g.a[0][0] = 1, g.a[1][0] = 0;
  if (x == 0) g.a[1][1] = 1, g.a[2][1] = 0;
  if (x == 1) g.a[2][2] = 1, g.a[3][2] = 0;
  if (x == 6) g.a[3][3] = g.a[4][4] = 1;
  if (x == 7) g.a[3][3] = 1, g.a[4][3] = 0;
  return g;
}
int n, q, l, r;
Mat t[N << 2], s;
char c[N];
void Make(int x, int l, int r) {
  if (l == r) {
    t[x] = Get(c[l] - '0');
    return;
  }
  Make((x << 1), l, ((l + r) >> 1)), Make((x << 1 | 1), ((l + r) >> 1) + 1, r),
      t[x] = t[(x << 1 | 1)] * t[(x << 1)];
}
Mat Query(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[x];
  Mat a(1), b(1);
  if (L <= ((l + r) >> 1)) a = Query((x << 1), l, ((l + r) >> 1), L, R);
  if (((l + r) >> 1) < R) b = Query((x << 1 | 1), ((l + r) >> 1) + 1, r, L, R);
  return b * a;
}
int main() {
  n = rd(), q = rd(), scanf("%s", c + 1);
  Make(1, 1, n);
  while (q--) {
    l = rd(), r = rd(), s = Mat(1);
    s = Query(1, 1, n, l, r),
    printf("%d\n", s.a[4][0] > (r - l + 1) ? -1 : s.a[4][0]);
  }
  return 0;
}

```