DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int64_t maxn = 1e5 + 10;
int64_t n;
struct node {
  int64_t v, minc, maxc, lazy, lazy2;
  node operator+(node const& t) const {
    return {v + t.v, min(minc, t.minc), max(maxc, t.maxc)};
  }
} tree[4 * maxn];
void flush(int64_t seg, int64_t ini, int64_t fim) {
  if (tree[seg].lazy == 0) return;
  assert(tree[seg].minc == tree[seg].maxc);
  tree[seg].v += tree[seg].lazy2 * (fim - ini + 1);
  tree[seg].minc = tree[seg].maxc = tree[seg].lazy;
  if (ini != fim) {
    tree[seg << 1].lazy = tree[seg].lazy;
    tree[seg << 1 | 1].lazy = tree[seg].lazy;
    tree[seg << 1].lazy2 += tree[seg].lazy2;
    tree[seg << 1 | 1].lazy2 += tree[seg].lazy2;
  }
  tree[seg].lazy = tree[seg].lazy2 = 0;
}
void build(int64_t seg = 1, int64_t ini = 1, int64_t fim = n) {
  if (ini == fim) {
    tree[seg].minc = tree[seg].maxc = ini;
    return;
  }
  int64_t meio = ini + fim >> 1;
  build(seg << 1, ini, meio);
  build(seg << 1 | 1, meio + 1, fim);
  tree[seg] = tree[seg << 1] + tree[seg << 1 | 1];
}
void update(int64_t l, int64_t r, int64_t v, int64_t seg = 1, int64_t ini = 1,
            int64_t fim = n) {
  flush(seg, ini, fim);
  if (fim < l || ini > r) return;
  if (l <= ini && fim <= r && tree[seg].minc == tree[seg].maxc) {
    tree[seg].lazy = v;
    tree[seg].lazy2 = abs(v - tree[seg].minc);
    flush(seg, ini, fim);
    return;
  }
  int64_t meio = ini + fim >> 1;
  update(l, r, v, seg << 1, ini, meio);
  update(l, r, v, seg << 1 | 1, meio + 1, fim);
  tree[seg] = tree[seg << 1] + tree[seg << 1 | 1];
}
int64_t query(int64_t l, int64_t r, int64_t seg = 1, int64_t ini = 1,
              int64_t fim = n) {
  flush(seg, ini, fim);
  if (fim < l || ini > r) return 0;
  if (l <= ini && fim <= r) return tree[seg].v;
  int64_t meio = ini + fim >> 1;
  int64_t p1 = query(l, r, seg << 1, ini, meio);
  int64_t p2 = query(l, r, seg << 1 | 1, meio + 1, fim);
  return p1 + p2;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int64_t m;
  cin >> n >> m;
  build();
  while (m--) {
    int64_t t;
    cin >> t;
    if (t == 1) {
      int64_t l, r, x;
      cin >> l >> r >> x;
      update(l, r, x);
    } else {
      int64_t l, r;
      cin >> l >> r;
      cout << query(l, r) << '\n';
    }
  }
  return 0;
}
