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;
struct node {
  int s[260][260];
} s[200];
int n, Q, a[100002], blk, lx[100002], rx[100002], bel[100002], cn,
    fa[2200000][2], js, ans[100002];
int tb[100002];
vector<int> b[200];
inline bool cmp(int x, int y) { return a[x] < a[y]; }
node solve(vector<int> v, int l, int r) {
  if (v.size() == 1) {
    node ret;
    ret.s[0][0] = v[0];
    return ret;
  }
  int mid = (l + r) >> 1;
  vector<int> v1, v2;
  node ret;
  int rev[260][2];
  for (int i = 0; i < v.size(); i++) {
    if (a[v[i]] <= mid)
      v1.push_back(v[i]), rev[i][0] = 1, rev[i][1] = v1.size() - 1;
    else
      v2.push_back(v[i]), rev[i][0] = 2, rev[i][1] = v2.size() - 1;
  }
  node r1 = solve(v1, l, mid), r2 = solve(v2, mid + 1, r);
  int len = v.size();
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      int s1 = -1, t1 = -1, s2 = -1, t2 = -1;
      for (int ii = i; ii <= j; ii++) {
        if (rev[ii][0] == 1) {
          if (s1 < 0) s1 = rev[ii][1];
          t1 = rev[ii][1];
        } else {
          if (s2 < 0) s2 = rev[ii][1];
          t2 = rev[ii][1];
        }
      }
      if (s1 < 0)
        ret.s[i][j] = r2.s[s2][t2];
      else if (s2 < 0)
        ret.s[i][j] = r1.s[s1][t1];
      else {
        ret.s[i][j] = ++js;
        fa[js][0] = r1.s[s1][t1], fa[js][1] = r2.s[s2][t2];
      }
    }
  return ret;
}
int main() {
  scanf("%d%d", &n, &Q);
  blk = sqrt(Q);
  js = n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (Q <= 500) {
    for (int i = 1; i <= Q; i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      for (int j = l; j <= r; j++) tb[j] = j;
      sort(tb + l, tb + r + 1, cmp);
      int la = tb[l];
      for (int j = l + 1; j <= r; j++) {
        fa[++js][0] = la;
        fa[js][1] = tb[j];
        la = js;
      }
      ans[i] = la;
    }
    printf("%d\n", js);
    for (int i = n + 1; i <= js; i++) printf("%d %d\n", fa[i][0], fa[i][1]);
    for (int i = 1; i <= Q; i++) printf("%d ", ans[i]);
    puts("");
    return 0;
  }
  for (int i = 1, j; i <= n; i = j + 1) {
    j = min(n, i + blk - 1);
    lx[++cn] = i, rx[cn] = j;
    for (int ii = i; ii <= j; ii++) bel[ii] = cn;
  }
  for (int i = 1; i <= cn; i++) {
    for (int j = 1; j <= n; j++)
      if (lx[i] <= a[j] && a[j] <= rx[i]) b[i].push_back(j);
    s[i] = solve(b[i], lx[i], rx[i]);
  }
  for (int i = 1; i <= Q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int la = 0;
    for (int j = 1; j <= cn; j++) {
      int pl = -1, pr = -1;
      for (int ii = 0; ii < b[j].size(); ii++)
        if (b[j][ii] >= l && b[j][ii] <= r) {
          pl = ii;
          break;
        }
      for (int ii = 0; ii < b[j].size(); ii++)
        if (b[j][ii] >= l && b[j][ii] <= r) pr = ii;
      if (pl == -1) continue;
      if (pr == -1) continue;
      if (pl > pr) {
        printf("%d %d %d %d\n", l, r, pl, pr);
        return 0;
      }
      if (!la) {
        la = s[j].s[pl][pr];
      } else {
        fa[++js][0] = la;
        fa[js][1] = s[j].s[pl][pr];
        la = js;
      }
    }
    ans[i] = la;
  }
  printf("%d\n", js);
  for (int i = n + 1; i <= js; i++) printf("%d %d\n", fa[i][0], fa[i][1]);
  for (int i = 1; i <= Q; i++) printf("%d ", ans[i]);
  puts("");
}