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>
struct MI {
 private:
  char bb[1 << 14];
  FILE *f;
  char *bs, *be;
  char e;
  bool o, l;

 public:
  MI() : f(stdin), bs(0), be(0) {}
  inline char get() {
    if (o) {
      o = 0;
      return e;
    }
    if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
    if (bs == be) {
      l = 1;
      return -1;
    };
    return *bs++;
  }
  inline void unget(char c) {
    o = 1;
    e = c;
  }
  template <class T>
  inline T read() {
    T r;
    *this > r;
    return r;
  }
  template <class T>
  inline MI &operator>(T &);
};
template <class T>
struct Q {
  const static bool U = T(-1) >= T(0);
  inline void operator()(MI &t, T &r) const {
    r = 0;
    char c;
    bool y = 0;
    if (U)
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (isdigit(c)) break;
      }
    else
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (c == '-') {
          c = t.get();
          if (isdigit(c)) {
            y = 1;
            break;
          };
        } else if (isdigit(c))
          break;
        ;
      };
    for (;;) {
      if (c == -1) goto E;
      if (isdigit(c))
        r = r * 10 + (c ^ 48);
      else
        break;
      c = t.get();
    }
    t.unget(c);
  E:;
    if (y) r = -r;
  }
};
template <>
struct Q<char> {};
template <class T>
inline MI &MI::operator>(T &t) {
  Q<T>()(*this, t);
  return *this;
}
template <class T>
std::ostream &operator<(std::ostream &out, const T &t) {
  return out << t;
}
using std::cout;
MI cin;
const int $n = 5005;
const int $m = (1 << 16) + 5;
template <class T>
using Vec = std::vector<T>;
struct range {
  uint16_t l, r;
};
inline range operator+(const range &A, const range &B) {
  return {std::min(A.l, B.l), std::max(A.r, B.r)};
}
int n, m, v[$n], f[$n][$n], g[$n][$n], ans[$m], tp;
Vec<int> pos[$n];
Vec<std::pair<int, int>> way;
Vec<range> e;
inline int midpoint(Vec<int> arr) {
  auto it = arr.begin() + (arr.size() - 1) / 2;
  std::nth_element(arr.begin(), it, arr.end(),
                   [](int x, int y) { return v[x] < v[y]; });
  return v[*it];
}
inline int merge(int x, int y) {
  if (e[x].l > e[y].r) std::swap(x, y);
  assert(e[x].r < e[y].l);
  way.emplace_back(x, y);
  e.push_back(e[x] + e[y]);
  return ++tp;
}
void build(const Vec<int> &pos) {
  const int n = pos.size();
  if (!n) return;
  if (n == 1) {
    const auto t = pos[0];
    f[t][t] = t;
    return;
  }
  const int mid = midpoint(pos);
  Vec<int> lo, hi, nxt(n, ::n + 1), pre(n, 0);
  auto col = [mid](int i) { return v[i] <= mid; };
  for (int i = 0, j = 0, lst = 0; i < n; ++i) {
    const auto p = pos[i];
    const auto w = col(p);
    if (w != lst) {
      while (j < i) nxt[j++] = p;
      lst = w;
    }
    (w ? lo : hi).push_back(p);
  }
  for (int i = n, j = n - 1, lst = 0; i--;) {
    const auto p = pos[i];
    const auto w = col(p);
    if (w != lst) {
      while (j > i) pre[j--] = p;
      lst = w;
    }
  }
  assert(std::abs<int>(lo.size() - hi.size()) <= 1);
  build(lo);
  build(hi);
  if (lo.empty() || hi.empty()) return;
  for (int o = n - 1; o; --o)
    for (int i = n - o; i--;) {
      const auto j = i + o, l = pos[i], r = pos[j], nl = nxt[i], nr = pre[j];
      if (nl > r || nr < l) continue;
      if (col(l) == col(r)) {
        assert(col(nl) == col(nr));
        f[l][r] = merge(f[l][r], f[nl][nr]);
      } else {
        assert(col(l) == col(nr) && col(nl) == col(r));
        f[l][r] = merge(f[l][nr], f[nl][r]);
      }
    }
}
int main() {
  cin > n > m;
  tp = n;
  const auto B = (int)std::sqrt(m), T = (n - 1) / B + 1;
  e.push_back({});
  for (int i = 1; i <= n; ++i) {
    pos[((v[i] = (cin.read<int>())) - 1) / B + 1].push_back(i);
    e.push_back({(uint16_t)v[i], (uint16_t)v[i]});
  }
  for (int i = 1; i <= T; ++i) build(pos[i]);
  for (int i = 1; i <= m; ++i) {
    const auto l = (cin.read<int>()), r = (cin.read<int>());
    if (g[l][r]) {
      ans[i] = g[l][r];
      continue;
    }
    int c = 0;
    for (int i = 1; i <= T; ++i) {
      auto lo = std::lower_bound(pos[i].begin(), pos[i].end(), l);
      auto hi = --std::upper_bound(pos[i].begin(), pos[i].end(), r);
      if (lo > hi) continue;
      const auto t = f[*lo][*hi];
      c = c ? merge(c, t) : t;
    }
    ans[i] = g[l][r] = c;
  }
  cout < tp < ('\n');
  for (auto [a, b] : way) cout < a < ' ' < b < ('\n');
  for (int i = 1; i <= m; ++i) cout < ans[i] < " \n"[i == m];
}