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, bl, cnt, ans[700100], a[10010], pos[10010], f[800][260][260];
vector<pair<int, int> > mg;
int merge(int x, int y) {
  if (!x || !y)
    return x | y;
  else
    return mg.push_back(make_pair(x, y)), ++cnt;
}
struct point {
  int l, r, g[260][260], p[4100];
  void solve(int x, int l, int r, vector<int> v) {
    if (l == r) return f[x][1][1] = pos[v[1]], void();
    int mid = (l + r) >> 1;
    vector<int> vl, vr;
    vl.push_back(0);
    vr.push_back(0);
    for (int i = 1; i < v.size(); i++)
      v[i] <= mid ? vl.push_back(v[i]) : vr.push_back(v[i]);
    solve(x << 1, l, mid, vl);
    solve(x << 1 | 1, mid + 1, r, vr);
    int l1 = 1, l2 = 0, r1 = 1, r2 = 0;
    for (int i = 1; i < v.size(); i++) {
      for (int j = i; j < v.size(); j++) {
        v[j] <= mid ? l2++ : r2++;
        f[x][i][j] = merge(f[x << 1][l1][l2], f[x << 1 | 1][r1][r2]);
      }
      v[i] <= mid ? l1++ : r1++;
      l2 = l1 - 1, r2 = r1 - 1;
    }
  }
  int query(int l, int r) { return g[p[l - 1] + 1][p[r]]; }
} t[65];
int main() {
  scanf("%d%d", &n, &q);
  bl = 256;
  cnt = n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  for (int i = 1; i <= (n - 1) / bl + 1; i++) {
    t[i].l = (i - 1) * bl + 1;
    t[i].r = min(n, i * bl);
    vector<int> v;
    v.push_back(0);
    for (int j = 1; j <= n; j++) {
      if (a[j] >= t[i].l && a[j] <= t[i].r) v.push_back(a[j]);
      t[i].p[j] = v.size() - 1;
    }
    t[i].solve(1, t[i].l, t[i].r, v);
    for (int j = 1; j < v.size(); j++)
      for (int k = j; k < v.size(); k++) t[i].g[j][k] = f[1][j][k];
  }
  for (int i = 1, l, r; i <= q; i++) {
    scanf("%d%d", &l, &r);
    for (int j = 1; j <= (n - 1) / bl + 1; j++)
      ans[i] = merge(ans[i], t[j].query(l, r));
  }
  printf("%d\n", cnt);
  for (int i = 0; i < mg.size(); i++)
    printf("%d %d\n", mg[i].first, mg[i].second);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  puts("");
}