#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 10;
const int MN = 2e5 + 10;
struct node {
  int a[4][4];
  node() { memset(a, 0, sizeof a); }
} seg[MN << 1];
int n, q;
vector<int> pos[12];
string s;
node merge(node A, node B) {
  node ret;
  for (int i = 0; i < 4; ++i)
    for (int j = i; j < 4; ++j) {
      ret.a[i][j] = INF;
      for (int l = i; l <= j; ++l)
        ret.a[i][j] = min(ret.a[i][j], A.a[i][l] + B.a[l][j]);
    }
  return ret;
}
node get(int l, int r) {
  node retL, retR;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (r & 1) retR = merge(seg[--r], retR);
    if (l & 1) retL = merge(retL, seg[l++]);
  }
  return merge(retL, retR);
}
int P(int d, int pl) {
  int tmp = lower_bound(pos[d].begin(), pos[d].end(), pl) - pos[d].begin() - 1;
  return (tmp < 0 ? -1 : pos[d][tmp]);
}
void input() {
  cin >> n >> q;
  cin >> s;
  for (int i = 0; i < n; ++i) {
    int x = s[i] - '0';
    if (x == 2) seg[i + n].a[0][0] = 1;
    if (x == 0) seg[i + n].a[1][1] = 1;
    if (x == 1) seg[i + n].a[2][2] = 1;
    if (x == 6) seg[i + n].a[3][3] = 1;
    pos[x].push_back(i);
  }
  for (int i = n - 1; i; --i) seg[i] = merge(seg[i << 1], seg[i << 1 | 1]);
}
void solve() {
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    int a7 = P(7, r);
    int a1 = P(1, a7);
    int a0 = P(0, a1);
    int a2 = P(2, a0);
    if (a2 < l) {
      cout << -1 << endl;
      continue;
    }
    node n1 = get(l, a2), n2 = get(a2 + 1, a0), n3 = get(a0 + 1, a1),
         n4 = get(a1 + 1, r);
    n1.a[0][0] = INF, n2.a[1][1] = INF, n3.a[2][2] = INF;
    node ans = merge(merge(merge(n1, n2), n3), n4);
    cout << ans.a[0][3] << endl;
  }
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  input();
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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 = 1e9 + 10;
const int MN = 2e5 + 10;
struct node {
  int a[4][4];
  node() { memset(a, 0, sizeof a); }
} seg[MN << 1];
int n, q;
vector<int> pos[12];
string s;
node merge(node A, node B) {
  node ret;
  for (int i = 0; i < 4; ++i)
    for (int j = i; j < 4; ++j) {
      ret.a[i][j] = INF;
      for (int l = i; l <= j; ++l)
        ret.a[i][j] = min(ret.a[i][j], A.a[i][l] + B.a[l][j]);
    }
  return ret;
}
node get(int l, int r) {
  node retL, retR;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (r & 1) retR = merge(seg[--r], retR);
    if (l & 1) retL = merge(retL, seg[l++]);
  }
  return merge(retL, retR);
}
int P(int d, int pl) {
  int tmp = lower_bound(pos[d].begin(), pos[d].end(), pl) - pos[d].begin() - 1;
  return (tmp < 0 ? -1 : pos[d][tmp]);
}
void input() {
  cin >> n >> q;
  cin >> s;
  for (int i = 0; i < n; ++i) {
    int x = s[i] - '0';
    if (x == 2) seg[i + n].a[0][0] = 1;
    if (x == 0) seg[i + n].a[1][1] = 1;
    if (x == 1) seg[i + n].a[2][2] = 1;
    if (x == 6) seg[i + n].a[3][3] = 1;
    pos[x].push_back(i);
  }
  for (int i = n - 1; i; --i) seg[i] = merge(seg[i << 1], seg[i << 1 | 1]);
}
void solve() {
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    int a7 = P(7, r);
    int a1 = P(1, a7);
    int a0 = P(0, a1);
    int a2 = P(2, a0);
    if (a2 < l) {
      cout << -1 << endl;
      continue;
    }
    node n1 = get(l, a2), n2 = get(a2 + 1, a0), n3 = get(a0 + 1, a1),
         n4 = get(a1 + 1, r);
    n1.a[0][0] = INF, n2.a[1][1] = INF, n3.a[2][2] = INF;
    node ans = merge(merge(merge(n1, n2), n3), n4);
    cout << ans.a[0][3] << endl;
  }
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  input();
  solve();
  return 0;
}

```