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 = (1 << 12) + 20;
const int S = (1 << 8) + 20;
const int Q = (1 << 16) + 20;
int n, q, a[N], B, tot, ret[Q], se;
vector<pair<int, int> > ans;
inline int bel(int x) { return (x - 1) / B + 1; }
inline int merge(int x, int y) {
  if (!x || !y) return x + y;
  ans.push_back(make_pair(x, y));
  return ++se;
}
struct block {
  int L, R, pos[N], cnt, f[S][S], t[S][S];
  inline void push(int x) { pos[++cnt] = x; }
  inline void build(int l, int r, vector<int> ve) {
    if (l == r) {
      f[ve[0]][ve[0]] = pos[ve[0]];
      return;
    }
    vector<int> ql, qr;
    int mid = (l + r) >> 1;
    for (int i : ve) {
      if (a[pos[i]] <= mid)
        ql.push_back(i);
      else
        qr.push_back(i);
    }
    int siz = (int)ve.size();
    build(l, mid, ql);
    build(mid + 1, r, qr);
    for (int i = 0; i < siz; ++i)
      for (int j = i; j < siz; ++j) t[ve[i]][ve[j]] = f[ve[i]][ve[j]];
    for (int i = 0; i < siz; ++i)
      for (int j = i; j < siz; ++j) {
        int pos1, pos2;
        int l1 = lower_bound(ql.begin(), ql.end(), ve[i]) - ql.begin();
        int r1 = upper_bound(ql.begin(), ql.end(), ve[j]) - ql.begin() - 1;
        if (l1 > r1)
          pos1 = 0;
        else
          pos1 = t[ql[l1]][ql[r1]];
        int l2 = lower_bound(qr.begin(), qr.end(), ve[i]) - qr.begin();
        int r2 = upper_bound(qr.begin(), qr.end(), ve[j]) - qr.begin() - 1;
        if (l2 > r2)
          pos2 = 0;
        else
          pos2 = t[qr[l2]][qr[r2]];
        f[ve[i]][ve[j]] = merge(pos1, pos2);
      }
  }
  inline void init(int x) {
    L = (x - 1) * B + 1;
    R = min(x * B, n);
    vector<int> ve;
    for (int i = 1; i <= cnt; ++i) ve.push_back(i);
    build(L, R, ve);
  }
  inline int query(int l, int r) {
    int ql = lower_bound(pos + 1, pos + cnt + 1, l) - pos;
    int qr = upper_bound(pos + 1, pos + cnt + 1, r) - pos - 1;
    if (ql > qr)
      return 0;
    else
      return f[ql][qr];
  }
} bl[(1 << 4) + 20];
int main() {
  scanf("%d%d", &n, &q);
  B = 256;
  tot = (n - 1) / B + 1;
  se = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    bl[bel(a[i])].push(i);
  }
  for (int i = 1; i <= tot; ++i) bl[i].init(i);
  for (int i = 1, l, r; i <= q; ++i) {
    scanf("%d%d", &l, &r);
    for (int j = 1; j <= tot; ++j) ret[i] = merge(ret[i], bl[j].query(l, r));
  }
  printf("%d\n", se);
  for (pair<int, int> u : ans) printf("%d %d\n", u.first, u.second);
  for (int i = 1; i <= q; ++i) printf("%d ", ret[i]);
  return 0;
}