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 const B = 1 << 8, C = 1 << 4 | 5, N = 1 << 12 | 5;
vector<int> v[1 << 15 | 3], id[1 << 15 | 3][B | 3];
int n, q, cnt, s, X[2200005], Y[2200005], L[C], R[C], tot, root[C], pos[N],
    bel[N], ls[1 << 15 | 3], rs[1 << 15 | 3], ans[1 << 16 | 5], a[N];
int query(int x, int l, int r) {
  int L = lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin(),
      R = upper_bound(v[x].begin(), v[x].end(), r) - v[x].begin();
  if (L == R) return 0;
  R--;
  return id[x][L][R];
}
int mer(int x, int y) {
  if (!x || !y) return x + y;
  s++;
  X[s] = x;
  Y[s] = y;
  return s;
}
void build(int l, int r, int &x) {
  if (!x) x = ++tot;
  if (l == r) {
    v[x].push_back(pos[l]);
    id[x][0].push_back(pos[l]);
    return;
  }
  int mid = (l + r) >> 1, len = r - l + 1;
  build(l, mid, ls[x]), build(mid + 1, r, rs[x]);
  v[x].resize(r - l + 1);
  merge(v[ls[x]].begin(), v[ls[x]].end(), v[rs[x]].begin(), v[rs[x]].end(),
        v[x].begin());
  for (int i = 0; i < len; i++) id[x][i].resize(len);
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++)
      id[x][i][j] =
          mer(query(ls[x], v[x][i], v[x][j]), query(rs[x], v[x][i], v[x][j]));
}
int main() {
  int l, r;
  scanf("%d%d", &n, &q);
  s = n;
  while (1) {
    cnt++;
    L[cnt] = (cnt - 1) * B + 1;
    R[cnt] = min(n, cnt * B);
    for (int i = L[cnt]; i <= R[cnt]; i++) bel[i] = cnt;
    if (R[cnt] == n) break;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  tot = cnt;
  for (int i = 1; i <= cnt; i++) build(L[i], R[i], root[i] = i);
  for (int _ = 1; _ <= q; _++) {
    scanf("%d%d", &l, &r);
    for (int i = 1; i <= cnt; i++) ans[_] = mer(ans[_], query(i, l, r));
  }
  printf("%d\n", s);
  for (int i = n + 1; i <= s; i++) printf("%d %d\n", X[i], Y[i]);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  return 0;
}