Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, q, a[5005], s = (1 << 8), l[100005], r[100005], b[5005], c[5005], k,
                   dp1[300][300][10], dp2[300][300][20], w[100005];
vector<pair<int, int> > ans;
void Gen(int p, int l, int r, int d) {
  if (l == r) {
    dp1[l][r][d] = b[p + l];
    return;
  }
  int mid = (l + r) / 2;
  Gen(p, l, mid, d + 1);
  Gen(p, mid + 1, r, d + 1);
  int x1 = 0, x2 = 0, y1, y2;
  while (x1 <= mid - l || x2 <= r - mid - 1) {
    y1 = 0, y2 = 0;
    while (x1 + y1 <= mid - l || x2 + y2 <= r - mid - 1) {
      if (x2 + y2 > r - mid - 1)
        dp1[l + x1 + x2][r - y1 - y2][d] = dp1[l + x1][mid - y1][d + 1];
      else if (x1 + y1 > mid - l)
        dp1[l + x1 + x2][r - y1 - y2][d] = dp1[mid + 1 + x2][r - y2][d + 1];
      else {
        ans.push_back(
            {dp1[l + x1][mid - y1][d + 1], dp1[mid + 1 + x2][r - y2][d + 1]});
        k++;
        dp1[l + x1 + x2][r - y1 - y2][d] = k;
      }
      if (x1 + y1 <= mid - l &&
          (x2 + y2 > r - mid - 1 || b[p + mid - y1] >= b[p + r - y2]))
        y1++;
      else
        y2++;
    }
    if (x1 <= mid - l &&
        (x2 > r - mid - 1 || b[p + l + x1] <= b[p + mid + 1 + x2]))
      c[p + l + x1 + x2] = b[p + l + x1], x1++;
    else
      c[p + l + x1 + x2] = b[p + mid + 1 + x2], x2++;
  }
  for (int i = p + l; i <= p + r; i++) b[i] = c[i];
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n >> q;
  k = n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  for (int i = 1; i <= n; i += s) {
    Gen(i, 0, min(n, i + s - 1) - i, 0);
    for (int j = 0; j <= min(n, i + s - 1) - i; j++) {
      for (int k = j; k <= min(n, i + s - 1) - i; k++) {
        dp2[j][k][i / s] = dp1[j][k][0];
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    int last = 0;
    for (int j = 1; j <= n; j += s) {
      int lt = j, rt = min(n, j + s - 1);
      if (b[lt] > r || b[rt] < l) continue;
      int first, second;
      first = lower_bound(b + lt, b + rt + 1, l) - (b + lt);
      second = upper_bound(b + lt, b + rt + 1, r) - (b + lt + 1);
      if (first > second) continue;
      if (last) {
        k++;
        ans.push_back({last, dp2[first][second][j / s]});
        last = k;
      } else {
        last = dp2[first][second][j / s];
      }
    }
    w[i] = last;
  }
  cout << n + ans.size() << '\n';
  for (pair<int, int> i : ans) cout << i.first << ' ' << i.second << '\n';
  for (int i = 1; i <= q; i++) cout << w[i] << ' ';
}