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;
struct Block {
  long long l, r, c;
  Block(long long l, long long r, long long c) : l(l), r(r), c(c) {}
};
bool operator<(Block a, Block b) {
  if (a.l != b.l) return a.l < b.l;
  if (a.r != b.r) return a.r < b.r;
  return a.c < b.c;
}
const int maxn = 1000000 + 10;
long long tr_a[maxn], tr_b[maxn];
void add(long long tr[maxn], long long at, long long what) {
  for (++at; at < maxn; at += -at & at) tr[at] += what;
}
void modify(long long l, long long r, long long c) {
  add(tr_a, l, c);
  add(tr_b, l, (l - 1) * c);
  add(tr_a, r + 1, -c);
  add(tr_b, r + 1, -r * c);
}
long long sum(long long tr[maxn], long long at) {
  long long s = 0;
  for (++at; at > 0; at -= -at & at) s += tr[at];
  return s;
}
long long query(long long l, long long r) {
  long long sr = r * sum(tr_a, r) - sum(tr_b, r);
  long long sl = (l - 1) * sum(tr_a, l - 1) - sum(tr_b, l - 1);
  return sr - sl;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset((tr_a), (0), sizeof(tr_a)), memset((tr_b), (0), sizeof(tr_b));
  int n, q;
  cin >> n >> q;
  set<Block> blocks;
  for (int _n(n + 1), i(1); i < _n; i++) blocks.insert(Block(i, i, i));
  for (int _n(q), _(0); _ < _n; _++) {
    long long op, l, r, x;
    cin >> op >> l >> r;
    if (op == 1) {
      cin >> x;
      auto it = blocks.upper_bound(Block(l, 1011111111, 1011111111));
      --it;
      while (it != blocks.end()) {
        if (it->l > r) break;
        Block tmp = *it;
        modify(max(l, tmp.l), min(r, tmp.r), labs(tmp.c - x));
        if (tmp.l < l) {
          blocks.insert(Block(tmp.l, l - 1, tmp.c));
        }
        if (tmp.r > r) {
          blocks.insert(Block(r + 1, tmp.r, tmp.c));
        }
        ++it;
        blocks.erase(blocks.find(tmp));
      }
      blocks.insert(Block(l, r, x));
    } else {
      long long ans = query(l, r);
      cout << ans << endl;
    }
  }
  return 0;
}
