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;
inline int read() {
  int first = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) {
    if (ch == '-') f = -1;
  }
  for (; isdigit(ch); ch = getchar()) {
    first = first * 10 + ch - 48;
  }
  return first * f;
}
const int mxN = 1 << 12;
const int mxM = 1 << 16;
const int mxB = 1 << 8;
int a[mxN + 3], ai[mxN + 3];
int n, m, siz, B;
int ans[mxM + 3];
vector<pair<int, int> > way;
int merge(int u, int v) {
  if (!u || !v) {
    return u + v;
  }
  way.push_back(make_pair(u, v));
  return ++siz;
}
int f[mxB * 2 + 3][mxB + 3][mxB + 3];
struct Block {
  int len, al, ar;
  void solve(int u, int l, int r, vector<int> vec) {
    if (l == r) {
      f[u][1][1] = ai[vec[1]];
      return;
    }
    int mid = (l + r) >> 1;
    vector<int> vecl(1), vecr(1);
    for (int i = 1; i < vec.size(); i++) {
      if (vec[i] <= mid) {
        vecl.push_back(vec[i]);
      } else {
        vecr.push_back(vec[i]);
      }
    }
    solve(u << 1, l, mid, vecl);
    solve(u << 1 | 1, mid + 1, r, vecr);
    int xl = 1, xr = 1;
    for (int i = 1; i < vec.size(); i++) {
      int yl = xl - 1, yr = xr - 1;
      for (int j = i; j < vec.size(); j++) {
        yl += (vec[j] <= mid), yr += (vec[j] > mid);
        f[u][i][j] = merge(f[u << 1][xl][yl], f[u << 1 | 1][xr][yr]);
      }
      xl += (vec[i] <= mid), xr += (vec[i] > mid);
    }
  }
  int id[mxN + 3], qwq[mxB + 3][mxB + 3];
  void build() {
    vector<int> vec(1);
    for (int i = 1; i <= n; i++) {
      if (a[i] >= al && a[i] <= ar)
        vec.push_back(a[i]), id[i] = id[i - 1] + 1;
      else
        id[i] = id[i - 1];
    }
    solve(1, al, ar, vec);
    for (int i = 1; i < vec.size(); i++)
      for (int j = 1; j < vec.size(); j++) qwq[i][j] = f[1][i][j];
  }
  int query(int l, int r) { return qwq[id[l - 1] + 1][id[r]]; }
} blo[18];
int main() {
  n = read(), m = read(), siz = n, B = min(n, mxB);
  for (int i = 1; i <= n; i++) a[i] = read(), ai[a[i]] = i;
  for (int i = 1; (i - 1) * B + 1 <= n; i++) {
    blo[i].al = (i - 1) * B + 1, blo[i].ar = i * B;
    blo[i].build();
  }
  for (int q = 1; q <= m; q++) {
    int l = read(), r = read();
    ans[q] = 0;
    for (int i = 1; i <= (n - 1) / B + 1; i++) {
      ans[q] = merge(ans[q], blo[i].query(l, r));
    }
  }
  printf("%d\n", siz);
  for (int i = 0; i < way.size(); i++)
    printf("%d %d\n", way[i].first, way[i].second);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}