#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 3;
char s[maxn];
struct T {
  int dp[5][5], l, r;
} t[maxn << 2];
T add(T a, T b, int l, int r) {
  T c;
  c.l = l, c.r = r;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      c.dp[i][j] = INF;
      for (int k = 0; k < 5; k++)
        c.dp[i][j] = min(c.dp[i][j], a.dp[i][k] + b.dp[k][j]);
    }
  return c;
}
void build(int p, int l, int r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) t[p].dp[i][j] = (i == j ? 0 : INF);
    if (s[l] == '2')
      t[p].dp[0][1] = 0, t[p].dp[0][0] = 1;
    else if (s[l] == '0')
      t[p].dp[1][2] = 0, t[p].dp[1][1] = 1;
    else if (s[l] == '1')
      t[p].dp[2][3] = 0, t[p].dp[2][2] = 1;
    else if (s[l] == '7')
      t[p].dp[3][4] = 0, t[p].dp[3][3] = 1;
    else if (s[l] == '6')
      t[p].dp[4][4] = t[p].dp[3][3] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  t[p] = add(t[p * 2], t[p * 2 + 1], t[p].l, t[p].r);
}
T query(int p, int l, int r) {
  if (l <= t[p].l && r >= t[p].r) return t[p];
  int mid = (t[p].l + t[p].r) >> 1;
  if (mid < l)
    return query(p * 2 + 1, l, r);
  else if (mid >= r)
    return query(p * 2, l, r);
  else
    return add(query(p * 2, l, r), query(p * 2 + 1, l, r), t[p].l, t[p].r);
}
int main() {
  int n, q;
  scanf("%d%d%s", &n, &q, s + 1);
  build(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    T ans = query(1, l, r);
    printf("%d\n", ans.dp[0][4] == INF ? -1 : ans.dp[0][4]);
  }
  return 0;
}

### Prompt

Please formulate 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;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 3;
char s[maxn];
struct T {
  int dp[5][5], l, r;
} t[maxn << 2];
T add(T a, T b, int l, int r) {
  T c;
  c.l = l, c.r = r;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      c.dp[i][j] = INF;
      for (int k = 0; k < 5; k++)
        c.dp[i][j] = min(c.dp[i][j], a.dp[i][k] + b.dp[k][j]);
    }
  return c;
}
void build(int p, int l, int r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) t[p].dp[i][j] = (i == j ? 0 : INF);
    if (s[l] == '2')
      t[p].dp[0][1] = 0, t[p].dp[0][0] = 1;
    else if (s[l] == '0')
      t[p].dp[1][2] = 0, t[p].dp[1][1] = 1;
    else if (s[l] == '1')
      t[p].dp[2][3] = 0, t[p].dp[2][2] = 1;
    else if (s[l] == '7')
      t[p].dp[3][4] = 0, t[p].dp[3][3] = 1;
    else if (s[l] == '6')
      t[p].dp[4][4] = t[p].dp[3][3] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  t[p] = add(t[p * 2], t[p * 2 + 1], t[p].l, t[p].r);
}
T query(int p, int l, int r) {
  if (l <= t[p].l && r >= t[p].r) return t[p];
  int mid = (t[p].l + t[p].r) >> 1;
  if (mid < l)
    return query(p * 2 + 1, l, r);
  else if (mid >= r)
    return query(p * 2, l, r);
  else
    return add(query(p * 2, l, r), query(p * 2 + 1, l, r), t[p].l, t[p].r);
}
int main() {
  int n, q;
  scanf("%d%d%s", &n, &q, s + 1);
  build(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    T ans = query(1, l, r);
    printf("%d\n", ans.dp[0][4] == INF ? -1 : ans.dp[0][4]);
  }
  return 0;
}

```