<p>You are given an integer <code>n</code> and a 2D integer array <code>queries</code>.</p>

<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code>. Initially, there is a <strong>unidirectional</strong> road from city <code>i</code> to city <code>i + 1</code> for all <code>0 &lt;= i &lt; n - 1</code>.</p>

<p><code>queries[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents the addition of a new <strong>unidirectional</strong> road from city <code>u<sub>i</sub></code> to city <code>v<sub>i</sub></code>. After each query, you need to find the <strong>length</strong> of the <strong>shortest path</strong> from city <code>0</code> to city <code>n - 1</code>.</p>

<p>There are no two queries such that <code>queries[i][0] &lt; queries[j][0] &lt; queries[i][1] &lt; queries[j][1]</code>.</p>

<p>Return an array <code>answer</code> where for each <code>i</code> in the range <code>[0, queries.length - 1]</code>, <code>answer[i]</code> is the <em>length of the shortest path</em> from city <code>0</code> to city <code>n - 1</code> after processing the <strong>first </strong><code>i + 1</code> queries.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, queries = [[2,4],[0,2],[0,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,2,1]</span></p>

<p><strong>Explanation: </strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image8.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image9.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image10.jpg" style="width: 350px; height: 96px;" /></p>

<p>After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4, queries = [[0,3],[0,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image11.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image12.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path remains 1.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= queries[i][0] &lt; queries[i][1] &lt; n</code></li>
	<li><code>1 &lt; queries[i][1] - queries[i][0]</code></li>
	<li>There are no repeated roads among the queries.</li>
	<li>There are no two queries such that <code>i != j</code> and <code>queries[i][0] &lt; queries[j][0] &lt; queries[i][1] &lt; queries[j][1]</code>.</li>
</ul>

import typing
class LazySegTree:
    def __init__(
            self,
            op: typing.Callable[[typing.Any, typing.Any], typing.Any],
            e: typing.Any,
            mapping: typing.Callable[[typing.Any, typing.Any], typing.Any],
            composition: typing.Callable[[typing.Any, typing.Any], typing.Any],
            id_: typing.Any,
            v: typing.Union[int, typing.List[typing.Any]]) -> None:
        self._op = op
        self._e = e
        self._mapping = mapping
        self._composition = composition
        self._id = id_

        if isinstance(v, int):
            v = [e] * v

        self._n = len(v)
        self._log = (self._n - 1).bit_length()
        self._size = 1 << self._log
        self._d = [e] * (2 * self._size)
        self._lz = [self._id] * self._size
        for i in range(self._n):
            self._d[self._size + i] = v[i]
        for i in range(self._size - 1, 0, -1):
            self._update(i)

    def set(self, p: int, x: typing.Any) -> None:
        assert 0 <= p < self._n

        p += self._size
        for i in range(self._log, 0, -1):
            self._push(p >> i)
        self._d[p] = x
        for i in range(1, self._log + 1):
            self._update(p >> i)

    def get(self, p: int) -> typing.Any:
        assert 0 <= p < self._n

        p += self._size
        for i in range(self._log, 0, -1):
            self._push(p >> i)
        return self._d[p]

    def prod(self, left: int, right: int) -> typing.Any:
        assert 0 <= left <= right <= self._n

        if left == right:
            return self._e

        left += self._size
        right += self._size

        for i in range(self._log, 0, -1):
            if ((left >> i) << i) != left:
                self._push(left >> i)
            if ((right >> i) << i) != right:
                self._push(right >> i)

        sml = self._e
        smr = self._e
        while left < right:
            if left & 1:
                sml = self._op(sml, self._d[left])
                left += 1
            if right & 1:
                right -= 1
                smr = self._op(self._d[right], smr)
            left >>= 1
            right >>= 1

        return self._op(sml, smr)

    def all_prod(self) -> typing.Any:
        return self._d[1]

    def apply(self, left: int, right: typing.Optional[int] = None,
              f: typing.Optional[typing.Any] = None) -> None:
        assert f is not None

        if right is None:
            p = left
            assert 0 <= left < self._n

            p += self._size
            for i in range(self._log, 0, -1):
                self._push(p >> i)
            self._d[p] = self._mapping(f, self._d[p])
            for i in range(1, self._log + 1):
                self._update(p >> i)
        else:
            assert 0 <= left <= right <= self._n
            if left == right:
                return

            left += self._size
            right += self._size

            for i in range(self._log, 0, -1):
                if ((left >> i) << i) != left:
                    self._push(left >> i)
                if ((right >> i) << i) != right:
                    self._push((right - 1) >> i)

            l2 = left
            r2 = right
            while left < right:
                if left & 1:
                    self._all_apply(left, f)
                    left += 1
                if right & 1:
                    right -= 1
                    self._all_apply(right, f)
                left >>= 1
                right >>= 1
            left = l2
            right = r2

            for i in range(1, self._log + 1):
                if ((left >> i) << i) != left:
                    self._update(left >> i)
                if ((right >> i) << i) != right:
                    self._update((right - 1) >> i)

    def max_right(
            self, left: int, g: typing.Callable[[typing.Any], bool]) -> int:
        assert 0 <= left <= self._n
        assert g(self._e)

        if left == self._n:
            return self._n

        left += self._size
        for i in range(self._log, 0, -1):
            self._push(left >> i)

        sm = self._e
        first = True
        while first or (left & -left) != left:
            first = False
            while left % 2 == 0:
                left >>= 1
            if not g(self._op(sm, self._d[left])):
                while left < self._size:
                    self._push(left)
                    left *= 2
                    if g(self._op(sm, self._d[left])):
                        sm = self._op(sm, self._d[left])
                        left += 1
                return left - self._size
            sm = self._op(sm, self._d[left])
            left += 1

        return self._n

    def min_left(self, right: int, g: typing.Any) -> int:
        assert 0 <= right <= self._n
        assert g(self._e)

        if right == 0:
            return 0

        right += self._size
        for i in range(self._log, 0, -1):
            self._push((right - 1) >> i)

        sm = self._e
        first = True
        while first or (right & -right) != right:
            first = False
            right -= 1
            while right > 1 and right % 2:
                right >>= 1
            if not g(self._op(self._d[right], sm)):
                while right < self._size:
                    self._push(right)
                    right = 2 * right + 1
                    if g(self._op(self._d[right], sm)):
                        sm = self._op(self._d[right], sm)
                        right -= 1
                return right + 1 - self._size
            sm = self._op(self._d[right], sm)

        return 0

    def _update(self, k: int) -> None:
        self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])

    def _all_apply(self, k: int, f: typing.Any) -> None:
        self._d[k] = self._mapping(f, self._d[k])
        if k < self._size:
            self._lz[k] = self._composition(f, self._lz[k])

    def _push(self, k: int) -> None:
        self._all_apply(2 * k, self._lz[k])
        self._all_apply(2 * k + 1, self._lz[k])
        self._lz[k] = self._id

def op(x, y):
    a1, b1 = divmod(x, 100001)
    a2, b2 = divmod(y, 100001)
    if a1 < a2: return x
    if a1 > a2: return y
    return a1 * (100001) + b1 + b2

ma = lambda x, y: x if x > y else y

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        seg = LazySegTree(op, 0, add, add, 0, [1] * n)
        outs = []
        for u, v in queries:
            seg.apply(u, v - 1, 100001)
            outs.append(seg.all_prod() % 100001 - 1)
        return outs
55400