Problem: One fine day Sasha went to the park for a walk. In the park, he saw that his favorite bench is occupied, and he had to sit down on the neighboring one. He sat down and began to listen to the silence. Suddenly, he got a question: what if in different parts of the park, the silence sounds in different ways? So it was. Let's divide the park into 1 × 1 meter squares and call them cells, and numerate rows from 1 to n from up to down, and columns from 1 to m from left to right. And now, every cell can be described with a pair of two integers (x, y), where x — the number of the row, and y — the number of the column. Sasha knows that the level of silence in the cell (i, j) equals to f_{i,j}, and all f_{i,j} form a permutation of numbers from 1 to n ⋅ m. Sasha decided to count, how many are there pleasant segments of silence?

Let's take some segment [l … r]. Denote S as the set of cells (i, j) that l ≤ f_{i,j} ≤ r. Then, the segment of silence [l … r] is pleasant if there is only one simple path between every pair of cells from S (path can't contain cells, which are not in S). In other words, set S should look like a tree on a plain. Sasha has done this task pretty quickly, and called the algorithm — "algorithm of silence's sounds".

Time passed, and the only thing left from the algorithm is a legend. To prove the truthfulness of this story, you have to help Sasha and to find the number of different pleasant segments of silence. Two segments [l_1 … r_1], [l_2 … r_2] are different, if l_1 ≠ l_2 or r_1 ≠ r_2 or both at the same time.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1000, 1 ≤ n ⋅ m ≤ 2 ⋅ 10^5) — the size of the park.

Each from next n lines contains m integers f_{i,j} (1 ≤ f_{i,j} ≤ n ⋅ m) — the level of silence in the cell with number (i, j).

It is guaranteed, that all f_{i,j} are different.

Output

Print one integer — the number of pleasant segments of silence.

Examples

Input

1 5
1 2 3 4 5


Output

15

Input

2 3
1 2 3
4 5 6


Output

15

Input

4 4
4 3 2 16
1 13 14 15
5 7 8 12
6 11 9 10


Output

50

Note

In the first example, all segments of silence are pleasant.

In the second example, pleasant segments of silence are the following:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct StVal {
  int minV = 0, numMin = 0;
  StVal() {}
  StVal(const int _v) : minV(_v), numMin(1) {}
  StVal(const StVal& v1, const StVal& v2) {
    if (v1.minV < v2.minV) {
      minV = v1.minV;
      numMin = v1.numMin;
    } else if (v1.minV > v2.minV) {
      minV = v2.minV;
      numMin = v2.numMin;
    } else {
      minV = v1.minV;
      numMin = v1.numMin + v2.numMin;
    }
  }
};
struct StUpdate {
  int v = 0;
  StUpdate() {}
  StUpdate(const int v) : v(v) {}
  StUpdate(const StUpdate& u1, const StUpdate& u2) { v = u1.v + u2.v; }
  void apply(StVal& sv, const int lo, const int hi) { sv.minV += v; }
};
struct SegTree {
  int n;
  vector<StVal> st;
  vector<StUpdate> lazy;
  vector<int> leaves;
  SegTree(const int n) : n(n) {
    leaves.resize(n);
    init(1, 0, n - 1);
    lazy.resize(st.size());
  }
  void init(const int si, const int lo, const int hi) {
    if (lo == hi) {
      if (si >= (int)st.size()) {
        st.resize(si + 1);
      }
      st[si] = StVal(0);
      leaves[lo] = si;
      return;
    }
    int lC = si << 1, rC = (si << 1) | 1, mid = (lo + hi) >> 1;
    init(lC, lo, mid);
    init(rC, mid + 1, hi);
    st[si] = StVal(st[lC], st[rC]);
  }
  void updateLazy(const int si, const int lo, const int hi) {
    lazy[si].apply(st[si], lo, hi);
    if (lo != hi) {
      lazy[si << 1] = StUpdate(lazy[si << 1], lazy[si]);
      lazy[si << 1 | 1] = StUpdate(lazy[si << 1 | 1], lazy[si]);
    }
    lazy[si] = StUpdate();
  }
  StVal query(const int l, const int r) {
    return (l <= r && l < n && r >= 0) ? query(l, r, 1, 0, n - 1) : StVal();
  }
  StVal query(const int l, const int r, const int si, const int lo,
              const int hi) {
    updateLazy(si, lo, hi);
    if (l <= lo && hi <= r) return st[si];
    const int mid = (lo + hi) >> 1;
    if (r <= mid) return query(l, r, si << 1, lo, mid);
    if (mid < l) return query(l, r, si << 1 | 1, mid + 1, hi);
    return StVal(query(l, r, si << 1, lo, mid),
                 query(l, r, si << 1 | 1, mid + 1, hi));
  }
  void update(const int l, const int r, const StUpdate u) {
    if (l <= r) update(l, r, u, 1, 0, n - 1);
  }
  void update(const int l, const int r, const StUpdate& u, const int si,
              const int lo, const int hi) {
    if (l <= lo && hi <= r) {
      lazy[si] = StUpdate(lazy[si], u);
      updateLazy(si, lo, hi);
    } else {
      updateLazy(si, lo, hi);
      if (hi < l || r < lo) return;
      const int mid = (lo + hi) >> 1;
      update(l, r, u, si << 1, lo, mid);
      update(l, r, u, si << 1 | 1, mid + 1, hi);
      st[si] = StVal(st[si << 1], st[si << 1 | 1]);
    }
  }
  void updateAllLazy() { updateAllLazy(1, 0, n - 1); }
  void updateAllLazy(const int si, const int lo, const int hi) {
    updateLazy(si, lo, hi);
    if (lo != hi) {
      const int mid = (lo + hi) >> 1;
      updateAllLazy(si << 1, lo, mid);
      updateAllLazy(si << 1 | 1, mid + 1, hi);
    }
  }
};
struct Node {
  array<Node*, 2> child{};
  Node* splayTreeParent = nullptr;
  Node* pathParentPointer = nullptr;
  bool reverse = false;
  Node() {}
  bool getSide() {
    return splayTreeParent ? splayTreeParent->child[1] == this : false;
  }
  void tryReverse() {
    if (!reverse) {
      return;
    }
    reverse = false;
    swap(child[0], child[1]);
    if (child[0]) {
      child[0]->reverse = !child[0]->reverse;
    }
    if (child[1]) {
      child[1]->reverse = !child[1]->reverse;
    }
  }
  Node* splay() {
    vector<Node*> ancestors;
    Node* cur = this;
    while (cur) {
      ancestors.push_back(cur);
      cur = cur->splayTreeParent;
    }
    for (int i = ancestors.size() - 1; i >= 0; i--) {
      if (!ancestors[i]) {
        continue;
      }
      ancestors[i]->tryReverse();
    }
    while (splayTreeParent) {
      if (splayTreeParent->splayTreeParent) {
        (getSide() == splayTreeParent->getSide() ? splayTreeParent : this)
            ->rotate();
      }
      rotate();
    }
    return this;
  }
  Node* rotate() {
    bool side = getSide(), parentSide = splayTreeParent->getSide();
    auto ancestor = splayTreeParent->splayTreeParent;
    pathParentPointer = splayTreeParent->pathParentPointer;
    splayTreeParent->pathParentPointer = nullptr;
    splayTreeParent->attach(child[!side], side);
    attach(splayTreeParent, !side);
    if (ancestor)
      ancestor->attach(this, parentSide);
    else
      splayTreeParent = nullptr;
    return this;
  }
  Node* attach(Node* node, int side) {
    if (node) node->splayTreeParent = this;
    child[side] = node;
    return this;
  }
};
struct LinkCutTreeGeneral {
  void link(Node* parent, Node* child) {
    assert(findRoot(child) != findRoot(parent));
    access(child);
    access(parent);
    Node* lChild = child->child[0];
    if (lChild) {
      lChild->reverse = !lChild->reverse;
      detachChild(child, 0);
    }
    child->attach(parent, 0);
  }
  void cut(Node* u) {
    access(u);
    assert(u->child[0]);
    u->child[0]->splayTreeParent = nullptr;
    u->child[0] = nullptr;
  }
  void cut(Node* u, Node* v) {
    bool b = findRoot(u) != findRoot(v);
    access(u);
    assert(!(b && u->child[0] && findMax(u->child[0]) == v));
    if (u->child[0] && findMax(u->child[0]) == v) {
      cut(u);
      return;
    }
    access(v);
    assert(!(b && v->child[0] && findMax(v->child[0]) == u));
    if (v->child[0] && findMax(v->child[0]) == u) {
      cut(v);
    }
  }
  Node* findRoot(Node* u) {
    access(u);
    while (u->child[0]) {
      u = u->child[0];
    }
    access(u);
    return u;
  }

 private:
  Node* access(Node* u) {
    u->splay();
    detachChild(u, 1);
    Node* curPP = u;
    while (u->pathParentPointer) {
      curPP = u->pathParentPointer;
      curPP->splay();
      detachChild(curPP, 1);
      curPP->attach(u, 1);
      u->pathParentPointer = nullptr;
      u->splay();
    }
    return curPP;
  }
  Node* findMax(Node* u) {
    if (!u) {
      return nullptr;
    }
    while (u->child[1]) {
      u = u->child[1];
    }
    return u;
  }
  void detachChild(Node* u, bool b) {
    if (!u->child[b]) {
      return;
    }
    u->child[b]->pathParentPointer = u;
    u->child[b]->splayTreeParent = nullptr;
    u->child[b] = nullptr;
  }
};
struct NodeWithId : Node {
  int id;
  NodeWithId(int id) : Node(), id(id) {}
};
struct DynamicTreeLCT {
  LinkCutTreeGeneral lct;
  vector<NodeWithId*> nodes;
  int n = 0;
  DynamicTreeLCT() {}
  int addNode() {
    nodes.push_back(new NodeWithId(n));
    return n++;
  }
  void link(int parentId, int childId) {
    lct.link(nodes[parentId], nodes[childId]);
  }
  void cut(int id) { lct.cut(nodes[id]); }
  void cut(int u, int v) { lct.cut(nodes[u], nodes[v]); }
  int findRoot(int u) {
    auto res = (NodeWithId*)lct.findRoot(nodes[u]);
    return res->id;
  }
  bool connected(int u, int v) { return findRoot(u) == findRoot(v); }
};
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
vector<int> numToID;
int n, m;
void cutNeighbors(DynamicTreeLCT& lct, vector<vector<int>>& grid, int l,
                  int r) {
  int i = numToID[l] / m, j = numToID[l] % m;
  int idx = numToID[l];
  for (int k = 0; k < 4; k++) {
    int ni = i + dx[k], nj = j + dy[k];
    if (ni < 0 || nj < 0 || ni == n || nj == m) {
      continue;
    }
    if (grid[ni][nj] < l || grid[ni][nj] > r) {
      continue;
    }
    int nIdx = numToID[grid[ni][nj]];
    lct.cut(idx, nIdx);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  DynamicTreeLCT lct;
  cin >> n >> m;
  vector<vector<int>> grid(n, vector<int>(m));
  numToID.resize(n * m + 1);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> grid[i][j];
      numToID[grid[i][j]] = lct.addNode();
    }
  }
  SegTree st(n * m + 1);
  long long ans = 0;
  int l = 1;
  for (int r = 1; r <= m * n; r++) {
    st.update(l, r, StUpdate(1));
    int i = numToID[r] / m, j = numToID[r] % m;
    int idx = numToID[r];
    for (int k = 0; k < 4; k++) {
      int ni = i + dx[k], nj = j + dy[k];
      if (ni < 0 || nj < 0 || ni == n || nj == m) {
        continue;
      }
      if (grid[ni][nj] < l || grid[ni][nj] > r) {
        continue;
      }
      int nIdx = numToID[grid[ni][nj]];
      while (lct.connected(idx, nIdx)) {
        cutNeighbors(lct, grid, l, r);
        l++;
      }
      if (l > grid[ni][nj]) {
        continue;
      }
      if (l == r) {
        break;
      }
      lct.link(idx, nIdx);
      st.update(l, grid[ni][nj], StUpdate(-1));
    }
    ans += st.query(l, r).numMin;
  }
  cout << ans << '\n';
}