#include <bits/stdc++.h>
using namespace std;
const int N = 2 * 1000 * 100 + 10, T = 5;
int n;
string s, t1 = "2016", t2 = "2017";
struct Node {
  int dp[T][T];
  Node() {
    for (int i = 0; i < T; i++)
      for (int j = 0; j < T; j++) dp[i][j] = (i ^ j ? N + 10 : 0);
  }
  void init(char c) {
    for (int i = 0; i < T; i++)
      for (int j = 0; j < T; j++) dp[i][j] = (i ^ j ? N + 10 : 0);
    for (int i = 0; i < 4; i++)
      if (c == t2[i]) dp[i][i + 1] = 0, dp[i][i] = 1;
    if (c == '6') dp[3][3] = dp[4][4] = 1;
  }
} seg[N << 2];
Node merge(Node L, Node R) {
  Node res = Node();
  for (int i = 0; i < T; i++) res.dp[i][i] = N + 10;
  for (int i = 0; i < T; i++)
    for (int j = i; j < T; j++)
      for (int k = i; k <= j; k++)
        res.dp[i][j] = min(res.dp[i][j], L.dp[i][k] + R.dp[k][j]);
  return res;
}
void build(int l = 0, int r = n, int root = 1) {
  if (l + 1 == r) {
    seg[root].init(s[l]);
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, root << 1), build(mid, r, root << 1 | 1);
  seg[root] = merge(seg[root << 1], seg[root << 1 | 1]);
}
Node get(int s, int e, int l = 0, int r = n, int root = 1) {
  if (s <= l && e >= r) return seg[root];
  if (s >= r || e <= l) return Node();
  int mid = l + r >> 1;
  return merge(get(s, e, l, mid, root << 1), get(s, e, mid, r, root << 1 | 1));
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int Q;
  cin >> n >> Q >> s;
  for (int i = 0; i < N << 2; i++) seg[i] = Node();
  build();
  while (Q--) {
    int l, r;
    cin >> l >> r;
    int res = get(--l, r).dp[0][4];
    cout << (res < N ? res : -1) << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 2 * 1000 * 100 + 10, T = 5;
int n;
string s, t1 = "2016", t2 = "2017";
struct Node {
  int dp[T][T];
  Node() {
    for (int i = 0; i < T; i++)
      for (int j = 0; j < T; j++) dp[i][j] = (i ^ j ? N + 10 : 0);
  }
  void init(char c) {
    for (int i = 0; i < T; i++)
      for (int j = 0; j < T; j++) dp[i][j] = (i ^ j ? N + 10 : 0);
    for (int i = 0; i < 4; i++)
      if (c == t2[i]) dp[i][i + 1] = 0, dp[i][i] = 1;
    if (c == '6') dp[3][3] = dp[4][4] = 1;
  }
} seg[N << 2];
Node merge(Node L, Node R) {
  Node res = Node();
  for (int i = 0; i < T; i++) res.dp[i][i] = N + 10;
  for (int i = 0; i < T; i++)
    for (int j = i; j < T; j++)
      for (int k = i; k <= j; k++)
        res.dp[i][j] = min(res.dp[i][j], L.dp[i][k] + R.dp[k][j]);
  return res;
}
void build(int l = 0, int r = n, int root = 1) {
  if (l + 1 == r) {
    seg[root].init(s[l]);
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, root << 1), build(mid, r, root << 1 | 1);
  seg[root] = merge(seg[root << 1], seg[root << 1 | 1]);
}
Node get(int s, int e, int l = 0, int r = n, int root = 1) {
  if (s <= l && e >= r) return seg[root];
  if (s >= r || e <= l) return Node();
  int mid = l + r >> 1;
  return merge(get(s, e, l, mid, root << 1), get(s, e, mid, r, root << 1 | 1));
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int Q;
  cin >> n >> Q >> s;
  for (int i = 0; i < N << 2; i++) seg[i] = Node();
  build();
  while (Q--) {
    int l, r;
    cin >> l >> r;
    int res = get(--l, r).dp[0][4];
    cout << (res < N ? res : -1) << '\n';
  }
  return 0;
}

```