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;
template <typename T>
inline void chkmin(T &a, const T &b) {
  a = a < b ? a : b;
}
template <typename T>
inline void chkmax(T &a, const T &b) {
  a = a > b ? a : b;
}
const int MAXN = 1 << 12, B = 1 << 8;
struct Data {
  int v, id;
} aa[MAXN];
vector<Data> ans;
int ids[MAXN], pla[1 << 16], n, Q, cnt;
vector<vector<int> > val[MAXN / B];
bool cmp1(const Data &a, const Data &b) { return a.v < b.v; }
bool cmp2(const Data &a, const Data &b) { return a.id < b.id; }
int push(int a, int b) {
  assert(++cnt <= 2200000);
  ans.push_back(Data{a, b});
  return cnt;
}
vector<vector<int> > divide(int l, int r) {
  if (l + 1 == r) return vector<vector<int> >(1, vector<int>(1, aa[l].id));
  int mid = (l + r) >> 1, pm = aa[mid].v;
  vector<vector<int> > vl = divide(l, mid), vr = divide(mid, r),
                       ans(r - l, vector<int>(r - l));
  sort(aa + l, aa + r, cmp2);
  int pl = 0, pr = 0;
  for (int i = l; i < r; i++) {
    int ql = pl, qr = pr;
    for (int j = i; j < r; j++) {
      ++(aa[j].v < pm ? ql : qr);
      int &cur = ans[i - l][j - l];
      if (pl == ql)
        cur = vr[pr][qr - 1];
      else if (pr == qr)
        cur = vl[pl][ql - 1];
      else
        cur = push(vl[pl][ql - 1], vr[pr][qr - 1]);
    }
    ++(aa[i].v < pm ? pl : pr);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &Q);
  cnt = n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &aa[i].v);
    aa[i].id = i + 1;
  }
  sort(aa, aa + n, cmp1);
  for (int i = 0; i < n; i += B) {
    int j = min(i + B, n);
    val[i / B] = divide(i, j);
  }
  for (int i = 0; i < n; i++) ids[i] = aa[i].id;
  for (int q = 0; q < Q; q++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int p = 0;
    for (int i = 0; i < n; i += B) {
      int j = min(i + B, n);
      int tl = lower_bound(ids + i, ids + j, l) - ids - i;
      int tr = upper_bound(ids + i, ids + j, r) - ids - 1 - i;
      int id = tl > tr ? 0 : val[i / B][tl][tr];
      if (id) p = p ? push(p, id) : id;
    }
    pla[q] = p;
  }
  printf("%d\n", cnt);
  for (const Data &d : ans) printf("%d %d\n", d.v, d.id);
  for (int i = 0; i < Q; i++) printf("%d ", pla[i]);
  return 0;
}