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 read() {
  int x = 0, pos = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') pos = 0;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
  return pos ? x : -x;
}
const int N = (1 << 16) + 400;
int cnt = 0, tot = 0;
int n, q;
int l[N], r[N];
int B, a[N], b[N], pls[N];
int rt[N], ls[N * 25], rs[N * 25];
map<unsigned long long, int> mp[N * 25];
pair<int, int> ans[3005200];
int tmp1[N], tmp2[N];
void solve(int &now, int l, int r) {
  if (!now) now = ++tot;
  if (l == r) {
    mp[now][(((unsigned long long)(l)) << 15) + l] = b[l];
    return;
  }
  int mid = (l + r) >> 1;
  int tp1 = 0, tp2 = 0;
  for (int i = l; i <= r; ++i) {
    if (a[b[i]] > mid)
      tmp2[++tp2] = b[i];
    else
      tmp1[++tp1] = b[i];
  }
  for (int i = 1; i <= tp1; ++i) {
    b[l + i - 1] = tmp1[i];
  }
  for (int i = 1; i <= tp2; ++i) {
    b[l + tp1 + i - 1] = tmp2[i];
  }
  solve(ls[now], l, mid);
  solve(rs[now], mid + 1, r);
  tp1 = l, tp2 = mid + 1;
  for (int i = l; i <= r; ++i) {
    if ((b[tp1] < b[tp2] && tp1 <= mid) || tp2 > r) {
      tmp1[i] = b[tp1++];
    } else
      tmp1[i] = b[tp2++];
  }
  for (int i = l; i <= r; ++i) b[i] = tmp1[i];
  int pl = l, pr = mid + 1;
  for (int i = l; i <= r; ++i) {
    int nr = pr, nl = pl;
    for (int j = i; j <= r; ++j) {
      if (a[b[j]] > mid)
        nr++;
      else
        nl++;
      if (nr == pr) {
        mp[now][(((unsigned long long)(i)) << 15) + j] =
            mp[ls[now]][(((unsigned long long)(pl)) << 15) + nl - 1];
      } else if (nl == pl) {
        mp[now][(((unsigned long long)(i)) << 15) + j] =
            mp[rs[now]][(((unsigned long long)(pr)) << 15) + nr - 1];
      } else {
        ans[++cnt] = {mp[ls[now]][(((unsigned long long)(pl)) << 15) + nl - 1],
                      mp[rs[now]][(((unsigned long long)(pr)) << 15) + nr - 1]};
        mp[now][(((unsigned long long)(i)) << 15) + j] = cnt;
      }
    }
    if (a[b[i]] > mid)
      pr++;
    else
      pl++;
  }
}
int stk[N];
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    b[a[i]] = i;
  }
  for (int i = 1; i <= q; ++i) {
    l[i] = read(), r[i] = read();
  }
  B = sqrt(q);
  int bl = n / B + ((n % B) > 0);
  cnt = n;
  for (int i = 1; i <= bl; ++i) {
    int l = 1 + (i - 1) * B, r = min(n, i * B);
    sort(b + l, b + r + 1);
    solve(rt[i], l, r);
  }
  for (int i = 1; i <= q; ++i) {
    int pr = 0, ncn = cnt;
    for (int j = 1; j <= bl; ++j) {
      int ql = 1 + (j - 1) * B, qr = min(n, j * B);
      int nl = lower_bound(b + ql, b + qr + 1, l[i]) - b;
      int nr = lower_bound(b + ql, b + qr + 1, r[i] + 1) - b;
      nr--;
      if (nl > nr) continue;
      if (!pr) {
        pr = mp[rt[j]][(((unsigned long long)(nl)) << 15) + nr];
        continue;
      } else {
        ans[++cnt] = {pr, mp[rt[j]][(((unsigned long long)(nl)) << 15) + nr]};
        pr = cnt;
      }
    }
    if (cnt != ncn)
      stk[i] = cnt;
    else
      stk[i] = pr;
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; ++i) {
    printf("%d %d\n", ans[i].first, ans[i].second);
  }
  for (int i = 1; i <= q; ++i) {
    printf("%d ", stk[i]);
  }
  return 0;
}