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;
const int N = 4100, M = 66000, lim = 2.2e6 + 1;
int a[N], ans[M], n, m, k, L[M], R[M], _l[N], _r[N], b[N], g[N], d[N];
int tmp[N], L0[N], R0[N], L1[N], R1[N];
int A[300][300], B[300][300];
int U[lim], V[lim];
inline int merge(int x, int y) {
  if (!x || !y) return x | y;
  ++k;
  U[k] = x, V[k] = y;
  return k;
}
void solve(int l, int r) {
  if (l == r) {
    int x = b[d[l]];
    A[x][x] = d[l];
    return;
  }
  const int mid = (l + r) / 2;
  solve(l, mid), solve(mid + 1, r);
  int i = l, j = mid + 1, t = 0;
  while (i <= mid && j <= r)
    if (d[i] < d[j])
      tmp[++t] = d[i++];
    else
      tmp[++t] = d[j++];
  while (i <= mid) tmp[++t] = d[i++];
  while (j <= r) tmp[++t] = d[j++];
  i = l, j = mid + 1;
  for (int x = 1; x <= t; ++x) {
    while (i <= mid && d[i] < tmp[x]) ++i;
    while (j <= r && d[j] < tmp[x]) ++j;
    L0[x] = i <= mid ? b[d[i]] : 299, L1[x] = j <= r ? b[d[j]] : 299;
  }
  i = mid, j = r;
  for (int x = t; x; --x) {
    while (i >= l && d[i] > tmp[x]) --i;
    while (j > mid && d[j] > tmp[x]) --j;
    R0[x] = i >= l ? b[d[i]] : 0, R1[x] = j > mid ? b[d[j]] : 0;
  }
  for (i = 1; i <= t; ++i)
    for (j = i; j <= t; ++j)
      B[b[tmp[i]]][b[tmp[j]]] = merge(A[L0[i]][R0[j]], A[L1[i]][R1[j]]);
  for (i = 1; i <= t; ++i)
    for (j = i; j <= t; ++j) A[b[tmp[i]]][b[tmp[j]]] = B[b[tmp[i]]][b[tmp[j]]];
  for (i = l; i <= r; ++i) d[i] = tmp[i - l + 1];
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m, k = n;
  for (int i = 1, x; i <= n; ++i) cin >> x, a[x] = g[x] = i;
  for (int i = 1; i <= m; ++i) cin >> L[i] >> R[i];
  for (int l = 1, r = 256; l <= n; l += 256, r += 256) {
    r = min(r, n);
    int len = r - l + 1;
    sort(g + l, g + r + 1);
    for (int i = l; i <= r; ++i) b[g[i]] = i - l + 1;
    for (int i = l; i <= r; ++i) d[i - l + 1] = a[i];
    solve(1, len);
    for (int i = 1, it = 1; i <= n; ++i) {
      while (it <= len && d[it] < i) ++it;
      _l[i] = it;
    }
    for (int i = n, it = len; i; --i) {
      while (it && d[it] > i) --it;
      _r[i] = it;
    }
    for (int i = 1; i <= m; ++i) ans[i] = merge(ans[i], A[_l[L[i]]][_r[R[i]]]);
  }
  cout << k << '\n';
  for (int i = n + 1; i <= k; ++i) cout << U[i] << ' ' << V[i] << '\n';
  for (int i = 1; i <= m; ++i) cout << ans[i] << " \n"[i == m];
  return 0;
}