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;
int N, M;
set<pair<long long, long long> > S;
class LazySegTree {
 private:
  pair<long long, long long> tree[2 * (1 << 18)];

 public:
  LazySegTree() {}
  void update(int l, int r, long long val, int id = 1, int low = 0,
              int high = (1 << 18)) {
    tree[id].first += (high - low) * tree[id].second;
    if (id < (1 << 18)) {
      tree[id * 2].second += tree[id].second;
      tree[id * 2 + 1].second += tree[id].second;
    }
    tree[id].second = 0;
    if (l <= low && r >= high) {
      tree[id].second += val;
      return;
    }
    tree[id].first += (min(r, high) - max(l, low)) * val;
    int mid = (low + high) / 2;
    if (r <= mid) return update(l, r, val, id * 2, low, mid);
    if (l >= mid) return update(l, r, val, id * 2 + 1, mid, high);
    update(l, r, val, id * 2, low, mid);
    update(l, r, val, id * 2 + 1, mid, high);
  }
  long long query(int l, int r, int id = 1, int low = 0, int high = (1 << 18)) {
    tree[id].first += (high - low) * tree[id].second;
    if (id < (1 << 18)) {
      tree[id * 2].second += tree[id].second;
      tree[id * 2 + 1].second += tree[id].second;
    }
    tree[id].second = 0;
    if (l <= low && r >= high) {
      return tree[id].first;
    }
    int mid = (low + high) / 2;
    if (r <= mid) return query(l, r, id * 2, low, mid);
    if (l >= mid) return query(l, r, id * 2 + 1, mid, high);
    return query(l, r, id * 2, low, mid) + query(l, r, id * 2 + 1, mid, high);
  }
};
LazySegTree tree;
void processType1(int l, int r, int x) {
  set<pair<long long, long long> >::iterator it =
      S.lower_bound(pair<long long, long long>(l, LLONG_MIN));
  pair<long long, long long> p = *it;
  if (p.second > 0) {
    S.insert(pair<long long, long long>(l - 1, p.second));
    S.insert(pair<long long, long long>(l, -p.second));
    it--;
  }
  while (it != S.end()) {
    pair<long long, long long> p1 = *it;
    it++;
    pair<long long, long long> p2 = *it;
    if (p1.first > r) {
      break;
    } else if (p2.first <= r) {
      tree.update(p1.first, p2.first + 1, abs(p2.second - x));
      it++;
      S.erase(prev(it));
      S.erase(prev(it));
    } else {
      tree.update(p1.first, r + 1, abs(p2.second - x));
      S.erase(prev(it));
      S.insert(pair<long long, long long>(r + 1, p1.second));
      break;
    }
  }
  S.insert(pair<long long, long long>(l, -x));
  S.insert(pair<long long, long long>(r, x));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M;
  for (int i = 1; i <= N; i++) {
    S.insert(pair<long long, long long>(i, -i));
    S.insert(pair<long long, long long>(i, i));
  }
  for (int i = 0; i < M; i++) {
    int type, l, r;
    long long x;
    cin >> type;
    if (type == 1) {
      cin >> l >> r >> x;
      processType1(l, r, x);
    } else {
      cin >> l >> r;
      cout << tree.query(l, r + 1) << endl;
    }
  }
}
